Deck 8: Linear Programming

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Question
Sketch the region described by the inequalities. Sketch the region described by the inequalities.    <div style=padding-top: 35px>
Sketch the region described by the inequalities.    <div style=padding-top: 35px>
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Question
x - y > -3 <strong>x - y > -3  </strong> A)   B)   C)   D)   <div style=padding-top: 35px>

A) <strong>x - y > -3  </strong> A)   B)   C)   D)   <div style=padding-top: 35px>
B) <strong>x - y > -3  </strong> A)   B)   C)   D)   <div style=padding-top: 35px>
C) <strong>x - y > -3  </strong> A)   B)   C)   D)   <div style=padding-top: 35px>
D) <strong>x - y > -3  </strong> A)   B)   C)   D)   <div style=padding-top: 35px>
Question
Maximize
Z = 5x - 3y
subject to
2x - y ≤ 8
2x - 5y ≥ 0
x - y = -2
x,y ≥ 0.
Question
To make some extra money,you make two types of picture frames,type A and type B,for sale.You have an initial start-up expense of $75.The production cost for type A is $3.60 per frame,and the production cost for type B is $5.20 per frame.The price for type A is $6.00 per frame and the price for type B is $10.00 per frame.Let x be the number of type A and y be the number of type B produced and sold.Write an inequality describing revenue less than cost.Solve the inequality and describe the region.Also,describe what this means in terms of frames.
Question
A car rental company has $540,000 to purchase up to 25 new cars of two different models.One model costs $18,000 each and the other model costs $24,000 each.Write a system of linear inequalities to describe the situation.Let x represent the first model and y represent the second.Find the region described by the system of linear inequalities.
Question
Sketch the region described by the inequalities. Sketch the region described by the inequalities.    <div style=padding-top: 35px>
Sketch the region described by the inequalities.    <div style=padding-top: 35px>
Question
4x + y > -2 <strong>4x + y > -2  </strong> A)   B)   C)   D)   <div style=padding-top: 35px>

A) <strong>4x + y > -2  </strong> A)   B)   C)   D)   <div style=padding-top: 35px>
B) <strong>4x + y > -2  </strong> A)   B)   C)   D)   <div style=padding-top: 35px>
C) <strong>4x + y > -2  </strong> A)   B)   C)   D)   <div style=padding-top: 35px>
D) <strong>4x + y > -2  </strong> A)   B)   C)   D)   <div style=padding-top: 35px>
Question
The XYZ Corporation produces two models of home computers,the Alpha model and the Beta model.Let x be the number of Alpha models and y the number of Beta models produced at the San Antonio factory per week.If the factory can produce at the most 100 Beta models in a week,write an inequality to describe this situation.Describe the region for the inequality.What additional inequalities can you add to the situation?
Question
The region indicated in the diagram <strong>The region indicated in the diagram   is described by</strong> A)   B)   C)   D)   E)   <div style=padding-top: 35px> is described by

A) <strong>The region indicated in the diagram   is described by</strong> A)   B)   C)   D)   E)   <div style=padding-top: 35px>
B) <strong>The region indicated in the diagram   is described by</strong> A)   B)   C)   D)   E)   <div style=padding-top: 35px>
C) <strong>The region indicated in the diagram   is described by</strong> A)   B)   C)   D)   E)   <div style=padding-top: 35px>
D) <strong>The region indicated in the diagram   is described by</strong> A)   B)   C)   D)   E)   <div style=padding-top: 35px>
E) <strong>The region indicated in the diagram   is described by</strong> A)   B)   C)   D)   E)   <div style=padding-top: 35px>
Question
Sketch the region described by the inequalities Sketch the region described by the inequalities    <div style=padding-top: 35px>
Sketch the region described by the inequalities    <div style=padding-top: 35px>
Question
4x - 2y ≤ 4 <strong>4x - 2y ≤ 4  </strong> A)   B)   C)   D)   <div style=padding-top: 35px>

A) <strong>4x - 2y ≤ 4  </strong> A)   B)   C)   D)   <div style=padding-top: 35px>
B) <strong>4x - 2y ≤ 4  </strong> A)   B)   C)   D)   <div style=padding-top: 35px>
C) <strong>4x - 2y ≤ 4  </strong> A)   B)   C)   D)   <div style=padding-top: 35px>
D) <strong>4x - 2y ≤ 4  </strong> A)   B)   C)   D)   <div style=padding-top: 35px>
Question
Maximize
Z = 4x + y
subject to
-x + y ≤ 2
3x + y ≤ 18
x,y ≥ 0.
Question
A chair company produces two models of chairs,the Sequoia and the Saratoga.The Sequoia model takes 3 hours to assemble and A chair company produces two models of chairs,the Sequoia and the Saratoga.The Sequoia model takes 3 hours to assemble and   hour to paint.The Saratoga model takes 2 hours to assemble and 1 hour to paint.The maximum number of hours available to assemble is 24 per day and the maximum number of hours available to paint is 8 per day. (a)If the company earns a profit of $20 per Sequoia model and $30 per Saratoga model,find the number of models produced per day in order to maximize profit. (b)If the company earns a profit of $30 per Sequoia model and $15 per Saratoga model,find the number of models produced per day in order to maximize profit. (c)Suppose the company decides to upgrade the two models so it takes an additional 2 hours to detail the Sequoia and 2 hours to detail the Saratoga.The maximum number of hours available to detail is 18 per day.If the company earns a profit of $45 per Sequoia model and $35 per Saratoga model,find the number of models produced per day in order to maximize profit. (d)Suppose the company decides to upgrade the two models so it takes an additional 2 hours to detail the Sequoia and 2 hours to detail the Saratoga.The maximum number of hours available to detail is 18 per day.If the company earns a profit of $30 per Sequoia model and $40 per Saratoga model,find the number of models produced per day in order to maximize profit.<div style=padding-top: 35px>
hour to paint.The Saratoga model takes 2 hours to assemble and 1 hour to paint.The maximum number of hours available to assemble is 24 per day and the maximum number of hours available to paint is 8 per day.
(a)If the company earns a profit of $20 per Sequoia model and $30 per Saratoga model,find the number of models produced per day in order to maximize profit.
(b)If the company earns a profit of $30 per Sequoia model and $15 per Saratoga model,find the number of models produced per day in order to maximize profit.
(c)Suppose the company decides to upgrade the two models so it takes an additional 2 hours to detail the Sequoia and 2 hours to detail the Saratoga.The maximum number of hours available to detail is 18 per day.If the company earns a profit of $45 per Sequoia model and $35 per Saratoga model,find the number of models produced per day in order to maximize profit.
(d)Suppose the company decides to upgrade the two models so it takes an additional 2 hours to detail the Sequoia and 2 hours to detail the Saratoga.The maximum number of hours available to detail is 18 per day.If the company earns a profit of $30 per Sequoia model and $40 per Saratoga model,find the number of models produced per day in order to maximize profit.
Question
The region indicated in the diagram <strong>The region indicated in the diagram   is described by</strong> A)   B)   C)   D)   E)   <div style=padding-top: 35px> is described by

A) <strong>The region indicated in the diagram   is described by</strong> A)   B)   C)   D)   E)   <div style=padding-top: 35px>
B) <strong>The region indicated in the diagram   is described by</strong> A)   B)   C)   D)   E)   <div style=padding-top: 35px>
C) <strong>The region indicated in the diagram   is described by</strong> A)   B)   C)   D)   E)   <div style=padding-top: 35px>
D) <strong>The region indicated in the diagram   is described by</strong> A)   B)   C)   D)   E)   <div style=padding-top: 35px>
E) <strong>The region indicated in the diagram   is described by</strong> A)   B)   C)   D)   E)   <div style=padding-top: 35px>
Question
Maximize
Z = 2x - 3y
subject to
2x + y ≥ 1
x - y ≤ 1
x,y ≥ 0.
Also find the corner point where the value of z is attained.
Question
Sketch the region described by the following system of inequalities: Sketch the region described by the following system of inequalities:    <div style=padding-top: 35px>
Sketch the region described by the following system of inequalities:    <div style=padding-top: 35px>
Question
Sketch the region described by the inequalities. Sketch the region described by the inequalities.    <div style=padding-top: 35px>
Sketch the region described by the inequalities.    <div style=padding-top: 35px>
Question
A store sells two types of calculators.In order to cover overhead,it must sell at least 40 calculators total per week,and in order to satisfy distribution requirements,it must sell at least twice as many of type II as type I.Write a system of inequalities to describe the situation.Let x be the number of type I that it sells in a week and y be the number of type II that it sells in a week.Find the region described by the system of linear inequalities.
Question
2x - 3y ≥ 6 <strong>2x - 3y ≥ 6  </strong> A)   B)   C)   D)   <div style=padding-top: 35px>

A) <strong>2x - 3y ≥ 6  </strong> A)   B)   C)   D)   <div style=padding-top: 35px>
B) <strong>2x - 3y ≥ 6  </strong> A)   B)   C)   D)   <div style=padding-top: 35px>
C) <strong>2x - 3y ≥ 6  </strong> A)   B)   C)   D)   <div style=padding-top: 35px>
D) <strong>2x - 3y ≥ 6  </strong> A)   B)   C)   D)   <div style=padding-top: 35px>
Question
x + y ≤ -2 <strong>x + y ≤ -2  </strong> A)   B)   C)   D)   <div style=padding-top: 35px>

A) <strong>x + y ≤ -2  </strong> A)   B)   C)   D)   <div style=padding-top: 35px>
B) <strong>x + y ≤ -2  </strong> A)   B)   C)   D)   <div style=padding-top: 35px>
C) <strong>x + y ≤ -2  </strong> A)   B)   C)   D)   <div style=padding-top: 35px>
D) <strong>x + y ≤ -2  </strong> A)   B)   C)   D)   <div style=padding-top: 35px>
Question
Maximize
Z = <strong>Maximize Z =   - 2   + 3   Subject to  </strong> A) 0 B) 5 C) 10 D) 15 E) 20 <div style=padding-top: 35px>
- 2 <strong>Maximize Z =   - 2   + 3   Subject to  </strong> A) 0 B) 5 C) 10 D) 15 E) 20 <div style=padding-top: 35px>
+ 3 <strong>Maximize Z =   - 2   + 3   Subject to  </strong> A) 0 B) 5 C) 10 D) 15 E) 20 <div style=padding-top: 35px>
Subject to <strong>Maximize Z =   - 2   + 3   Subject to  </strong> A) 0 B) 5 C) 10 D) 15 E) 20 <div style=padding-top: 35px>

A) 0
B) 5
C) 10
D) 15
E) 20
Question
When no quotients exist in a simplex table,

A) a degenerate BFS will occur.
B) the problem has an unbounded solution.
C) there is a possibility for multiple optimum solutions.
Question
Use the simplex method to minimize
Z = 3 <strong>Use the simplex method to minimize Z = 3   + 2   Subject to       ,   ≥ 0</strong> A) Z =   when   =   ,   = 0 B) Z = 21 when   = 7,   = 0 C) Z = 14 when   = 0,   = 7 D) Z = 14 when   = 7,   = 0 <div style=padding-top: 35px>
+ 2 <strong>Use the simplex method to minimize Z = 3   + 2   Subject to       ,   ≥ 0</strong> A) Z =   when   =   ,   = 0 B) Z = 21 when   = 7,   = 0 C) Z = 14 when   = 0,   = 7 D) Z = 14 when   = 7,   = 0 <div style=padding-top: 35px>
Subject to <strong>Use the simplex method to minimize Z = 3   + 2   Subject to       ,   ≥ 0</strong> A) Z =   when   =   ,   = 0 B) Z = 21 when   = 7,   = 0 C) Z = 14 when   = 0,   = 7 D) Z = 14 when   = 7,   = 0 <div style=padding-top: 35px>
<strong>Use the simplex method to minimize Z = 3   + 2   Subject to       ,   ≥ 0</strong> A) Z =   when   =   ,   = 0 B) Z = 21 when   = 7,   = 0 C) Z = 14 when   = 0,   = 7 D) Z = 14 when   = 7,   = 0 <div style=padding-top: 35px>
<strong>Use the simplex method to minimize Z = 3   + 2   Subject to       ,   ≥ 0</strong> A) Z =   when   =   ,   = 0 B) Z = 21 when   = 7,   = 0 C) Z = 14 when   = 0,   = 7 D) Z = 14 when   = 7,   = 0 <div style=padding-top: 35px>
, <strong>Use the simplex method to minimize Z = 3   + 2   Subject to       ,   ≥ 0</strong> A) Z =   when   =   ,   = 0 B) Z = 21 when   = 7,   = 0 C) Z = 14 when   = 0,   = 7 D) Z = 14 when   = 7,   = 0 <div style=padding-top: 35px>
≥ 0

A) Z = <strong>Use the simplex method to minimize Z = 3   + 2   Subject to       ,   ≥ 0</strong> A) Z =   when   =   ,   = 0 B) Z = 21 when   = 7,   = 0 C) Z = 14 when   = 0,   = 7 D) Z = 14 when   = 7,   = 0 <div style=padding-top: 35px> when <strong>Use the simplex method to minimize Z = 3   + 2   Subject to       ,   ≥ 0</strong> A) Z =   when   =   ,   = 0 B) Z = 21 when   = 7,   = 0 C) Z = 14 when   = 0,   = 7 D) Z = 14 when   = 7,   = 0 <div style=padding-top: 35px> = <strong>Use the simplex method to minimize Z = 3   + 2   Subject to       ,   ≥ 0</strong> A) Z =   when   =   ,   = 0 B) Z = 21 when   = 7,   = 0 C) Z = 14 when   = 0,   = 7 D) Z = 14 when   = 7,   = 0 <div style=padding-top: 35px> , <strong>Use the simplex method to minimize Z = 3   + 2   Subject to       ,   ≥ 0</strong> A) Z =   when   =   ,   = 0 B) Z = 21 when   = 7,   = 0 C) Z = 14 when   = 0,   = 7 D) Z = 14 when   = 7,   = 0 <div style=padding-top: 35px> = 0
B) Z = 21 when <strong>Use the simplex method to minimize Z = 3   + 2   Subject to       ,   ≥ 0</strong> A) Z =   when   =   ,   = 0 B) Z = 21 when   = 7,   = 0 C) Z = 14 when   = 0,   = 7 D) Z = 14 when   = 7,   = 0 <div style=padding-top: 35px> = 7, <strong>Use the simplex method to minimize Z = 3   + 2   Subject to       ,   ≥ 0</strong> A) Z =   when   =   ,   = 0 B) Z = 21 when   = 7,   = 0 C) Z = 14 when   = 0,   = 7 D) Z = 14 when   = 7,   = 0 <div style=padding-top: 35px> = 0
C) Z = 14 when <strong>Use the simplex method to minimize Z = 3   + 2   Subject to       ,   ≥ 0</strong> A) Z =   when   =   ,   = 0 B) Z = 21 when   = 7,   = 0 C) Z = 14 when   = 0,   = 7 D) Z = 14 when   = 7,   = 0 <div style=padding-top: 35px> = 0, <strong>Use the simplex method to minimize Z = 3   + 2   Subject to       ,   ≥ 0</strong> A) Z =   when   =   ,   = 0 B) Z = 21 when   = 7,   = 0 C) Z = 14 when   = 0,   = 7 D) Z = 14 when   = 7,   = 0 <div style=padding-top: 35px> = 7
D) Z = 14 when <strong>Use the simplex method to minimize Z = 3   + 2   Subject to       ,   ≥ 0</strong> A) Z =   when   =   ,   = 0 B) Z = 21 when   = 7,   = 0 C) Z = 14 when   = 0,   = 7 D) Z = 14 when   = 7,   = 0 <div style=padding-top: 35px> = 7, <strong>Use the simplex method to minimize Z = 3   + 2   Subject to       ,   ≥ 0</strong> A) Z =   when   =   ,   = 0 B) Z = 21 when   = 7,   = 0 C) Z = 14 when   = 0,   = 7 D) Z = 14 when   = 7,   = 0 <div style=padding-top: 35px> = 0
Question
Maximize
Z = 10x + 15y
subject to Maximize Z = 10x + 15y subject to  <div style=padding-top: 35px>
Question
A manufacturer produces two products,product A and product B.Both products require processing on Machines I and II.The number of hours needed to produce one unit is given by the following chart: A manufacturer produces two products,product A and product B.Both products require processing on Machines I and II.The number of hours needed to produce one unit is given by the following chart:   Machine I is available for at most 1150 hours and Machine II is available for at most 1100 hours.If the profit made on product A is $15 / unit and the profit made on product B is $30 / unit.Find the production level that will maximize profit and find the Maximum profit.<div style=padding-top: 35px>
Machine I is available for at most 1150 hours and Machine II is available for at most 1100 hours.If the profit made on product A is $15 / unit and the profit made on product B is $30 / unit.Find the production level that will maximize profit and find the Maximum profit.
Question
A person decides to take two different dietary supplements.Each supplement contains two essential ingredients,A and B,for which there are minimum daily requirements,and each contains a third ingredient,C,which needs to be minimized.Find the amount of each supplement that the person should take each day in order to satisfy the requirements for A and B while minimizing C.Also find the amount of C the person takes each day. A person decides to take two different dietary supplements.Each supplement contains two essential ingredients,A and B,for which there are minimum daily requirements,and each contains a third ingredient,C,which needs to be minimized.Find the amount of each supplement that the person should take each day in order to satisfy the requirements for A and B while minimizing C.Also find the amount of C the person takes each day.  <div style=padding-top: 35px>
Question
Use the simplex method to
maximize
Z = Use the simplex method to maximize Z =   + 4   +   subject to   .<div style=padding-top: 35px>
+ 4 Use the simplex method to maximize Z =   + 4   +   subject to   .<div style=padding-top: 35px>
+ Use the simplex method to maximize Z =   + 4   +   subject to   .<div style=padding-top: 35px>
subject to Use the simplex method to maximize Z =   + 4   +   subject to   .<div style=padding-top: 35px>
.
Question
A summer camp wants to hire counselors and aides to fill its staffing needs at minimum cost.The average monthly salary of a counselor is $2400 and the average monthly salary of an aide is $1100.The camp can accommodate up to 45 staff members and needs at least 30 to run properly.They must have at least 10 aides,and may have up to 3 aides for every 2 counselors.How many counselors and how many aides should the camp hire to minimize cost?

A) 12 counselors and 18 aides
B) 27 counselors and 18 aides
C) 35 counselors and 10 aides
D) 18 counselors and 12 aides
Question
A company has two different locations to assemble three different models of PCs.The table below summarizes the daily production capacity,the minimum number of each type needed,and the daily operating costs for each location.Find the number of days that each location needs to operate in order to fill the orders at minimum cost. A company has two different locations to assemble three different models of PCs.The table below summarizes the daily production capacity,the minimum number of each type needed,and the daily operating costs for each location.Find the number of days that each location needs to operate in order to fill the orders at minimum cost.  <div style=padding-top: 35px>
Question
Give the dual of:
Maximize
Z = Give the dual of: Maximize Z =   + 3   + 4   subject to 3   +   -   ≤ 4 4   -   + 2   ≥ 1   ,   ,   ≥ 0.<div style=padding-top: 35px>
+ 3 Give the dual of: Maximize Z =   + 3   + 4   subject to 3   +   -   ≤ 4 4   -   + 2   ≥ 1   ,   ,   ≥ 0.<div style=padding-top: 35px>
+ 4 Give the dual of: Maximize Z =   + 3   + 4   subject to 3   +   -   ≤ 4 4   -   + 2   ≥ 1   ,   ,   ≥ 0.<div style=padding-top: 35px>
subject to
3 Give the dual of: Maximize Z =   + 3   + 4   subject to 3   +   -   ≤ 4 4   -   + 2   ≥ 1   ,   ,   ≥ 0.<div style=padding-top: 35px>
+ Give the dual of: Maximize Z =   + 3   + 4   subject to 3   +   -   ≤ 4 4   -   + 2   ≥ 1   ,   ,   ≥ 0.<div style=padding-top: 35px>
- Give the dual of: Maximize Z =   + 3   + 4   subject to 3   +   -   ≤ 4 4   -   + 2   ≥ 1   ,   ,   ≥ 0.<div style=padding-top: 35px>
≤ 4
4 Give the dual of: Maximize Z =   + 3   + 4   subject to 3   +   -   ≤ 4 4   -   + 2   ≥ 1   ,   ,   ≥ 0.<div style=padding-top: 35px>
- Give the dual of: Maximize Z =   + 3   + 4   subject to 3   +   -   ≤ 4 4   -   + 2   ≥ 1   ,   ,   ≥ 0.<div style=padding-top: 35px>
+ 2 Give the dual of: Maximize Z =   + 3   + 4   subject to 3   +   -   ≤ 4 4   -   + 2   ≥ 1   ,   ,   ≥ 0.<div style=padding-top: 35px>
≥ 1 Give the dual of: Maximize Z =   + 3   + 4   subject to 3   +   -   ≤ 4 4   -   + 2   ≥ 1   ,   ,   ≥ 0.<div style=padding-top: 35px>
, Give the dual of: Maximize Z =   + 3   + 4   subject to 3   +   -   ≤ 4 4   -   + 2   ≥ 1   ,   ,   ≥ 0.<div style=padding-top: 35px>
, Give the dual of: Maximize Z =   + 3   + 4   subject to 3   +   -   ≤ 4 4   -   + 2   ≥ 1   ,   ,   ≥ 0.<div style=padding-top: 35px>
≥ 0.
Question
In the initial simplex table below,find the pivot entry. <strong>In the initial simplex table below,find the pivot entry.      </strong> A) -1 B) 0 C) 10 D) 2 E) 6 <div style=padding-top: 35px> <strong>In the initial simplex table below,find the pivot entry.      </strong> A) -1 B) 0 C) 10 D) 2 E) 6 <div style=padding-top: 35px>
<strong>In the initial simplex table below,find the pivot entry.      </strong> A) -1 B) 0 C) 10 D) 2 E) 6 <div style=padding-top: 35px>

A) -1
B) 0
C) 10
D) 2
E) 6
Question
In the initial simplex table below,find the departing variable. <strong>In the initial simplex table below,find the departing variable.      </strong> A)   B)   C)   D)   E) Z <div style=padding-top: 35px> <strong>In the initial simplex table below,find the departing variable.      </strong> A)   B)   C)   D)   E) Z <div style=padding-top: 35px>
<strong>In the initial simplex table below,find the departing variable.      </strong> A)   B)   C)   D)   E) Z <div style=padding-top: 35px>

A) <strong>In the initial simplex table below,find the departing variable.      </strong> A)   B)   C)   D)   E) Z <div style=padding-top: 35px>
B) <strong>In the initial simplex table below,find the departing variable.      </strong> A)   B)   C)   D)   E) Z <div style=padding-top: 35px>
C) <strong>In the initial simplex table below,find the departing variable.      </strong> A)   B)   C)   D)   E) Z <div style=padding-top: 35px>
D) <strong>In the initial simplex table below,find the departing variable.      </strong> A)   B)   C)   D)   E) Z <div style=padding-top: 35px>
E) Z
Question
A company manufactures two models of inline skates,Alpha and Beta,at two different manufacturing plants.The maximum output at plant I is 1000 per month,while the maximum output at plant II is 1200 per month.Due to contractual obligations,the number of Alpha models produced at plant I must exceed the number of Beta models produced at plant I by at least 100.The profit per Alpha and Beta model manufactured at plant I is $50 and $80,respectively,while the profit per Alpha and Beta model manufactured at plant II is $60 and $70,respectively.This month,the company received an order for 900 Alpha and 1000 Beta models.Find how many of each model should be produced at each plant in order to satisfy the order and maximize the profit.(Hint: Let A company manufactures two models of inline skates,Alpha and Beta,at two different manufacturing plants.The maximum output at plant I is 1000 per month,while the maximum output at plant II is 1200 per month.Due to contractual obligations,the number of Alpha models produced at plant I must exceed the number of Beta models produced at plant I by at least 100.The profit per Alpha and Beta model manufactured at plant I is $50 and $80,respectively,while the profit per Alpha and Beta model manufactured at plant II is $60 and $70,respectively.This month,the company received an order for 900 Alpha and 1000 Beta models.Find how many of each model should be produced at each plant in order to satisfy the order and maximize the profit.(Hint: Let   represent the number of Alpha models and   represent the number of Beta models manufactured at plant I.)<div style=padding-top: 35px>
represent the number of Alpha models and A company manufactures two models of inline skates,Alpha and Beta,at two different manufacturing plants.The maximum output at plant I is 1000 per month,while the maximum output at plant II is 1200 per month.Due to contractual obligations,the number of Alpha models produced at plant I must exceed the number of Beta models produced at plant I by at least 100.The profit per Alpha and Beta model manufactured at plant I is $50 and $80,respectively,while the profit per Alpha and Beta model manufactured at plant II is $60 and $70,respectively.This month,the company received an order for 900 Alpha and 1000 Beta models.Find how many of each model should be produced at each plant in order to satisfy the order and maximize the profit.(Hint: Let   represent the number of Alpha models and   represent the number of Beta models manufactured at plant I.)<div style=padding-top: 35px>
represent the number of Beta models manufactured at plant I.)
Question
A manufacturer produces two products,product A and product B.Both products require processing on Machines I and II.The number of hours needed to produce one unit is given by the following chart: A manufacturer produces two products,product A and product B.Both products require processing on Machines I and II.The number of hours needed to produce one unit is given by the following chart:   Machine I is available for at most 1000 hours and Machine II is available for at most 2500 hours.If the profit made on product A is $20 / unit and the profit made on product B is $25 / unit.Find the production level that will maximize profit and find the Maximum profit.<div style=padding-top: 35px>
Machine I is available for at most 1000 hours and Machine II is available for at most 2500 hours.If the profit made on product A is $20 / unit and the profit made on product B is $25 / unit.Find the production level that will maximize profit and find the Maximum profit.
Question
Use the simplex method to minimize
Z = 4 Use the simplex method to minimize Z = 4   +   subject to   +   ≥ 5   + 2   ≥ 8   ,   ≥ 0.<div style=padding-top: 35px>
+ Use the simplex method to minimize Z = 4   +   subject to   +   ≥ 5   + 2   ≥ 8   ,   ≥ 0.<div style=padding-top: 35px>
subject to Use the simplex method to minimize Z = 4   +   subject to   +   ≥ 5   + 2   ≥ 8   ,   ≥ 0.<div style=padding-top: 35px>
+ Use the simplex method to minimize Z = 4   +   subject to   +   ≥ 5   + 2   ≥ 8   ,   ≥ 0.<div style=padding-top: 35px>
≥ 5 Use the simplex method to minimize Z = 4   +   subject to   +   ≥ 5   + 2   ≥ 8   ,   ≥ 0.<div style=padding-top: 35px>
+ 2 Use the simplex method to minimize Z = 4   +   subject to   +   ≥ 5   + 2   ≥ 8   ,   ≥ 0.<div style=padding-top: 35px>
≥ 8 Use the simplex method to minimize Z = 4   +   subject to   +   ≥ 5   + 2   ≥ 8   ,   ≥ 0.<div style=padding-top: 35px>
, Use the simplex method to minimize Z = 4   +   subject to   +   ≥ 5   + 2   ≥ 8   ,   ≥ 0.<div style=padding-top: 35px>
≥ 0.
Question
Maximize
Z = 4x + 6y
subject to Maximize Z = 4x + 6y subject to  <div style=padding-top: 35px>
Question
Use the simplex method to
maximize
Z = 2 Use the simplex method to maximize Z = 2   - 3   subject to   +   ≥ 5   + 2   ≤ 8   ,   ≥ 0<div style=padding-top: 35px>
- 3 Use the simplex method to maximize Z = 2   - 3   subject to   +   ≥ 5   + 2   ≤ 8   ,   ≥ 0<div style=padding-top: 35px>
subject to Use the simplex method to maximize Z = 2   - 3   subject to   +   ≥ 5   + 2   ≤ 8   ,   ≥ 0<div style=padding-top: 35px>
+ Use the simplex method to maximize Z = 2   - 3   subject to   +   ≥ 5   + 2   ≤ 8   ,   ≥ 0<div style=padding-top: 35px>
≥ 5 Use the simplex method to maximize Z = 2   - 3   subject to   +   ≥ 5   + 2   ≤ 8   ,   ≥ 0<div style=padding-top: 35px>
+ 2 Use the simplex method to maximize Z = 2   - 3   subject to   +   ≥ 5   + 2   ≤ 8   ,   ≥ 0<div style=padding-top: 35px>
≤ 8 Use the simplex method to maximize Z = 2   - 3   subject to   +   ≥ 5   + 2   ≤ 8   ,   ≥ 0<div style=padding-top: 35px>
, Use the simplex method to maximize Z = 2   - 3   subject to   +   ≥ 5   + 2   ≤ 8   ,   ≥ 0<div style=padding-top: 35px>
≥ 0
Question
Use the simplex method to solve the following problem: United Blimpo Co.produces two types of exercise devices,regular and heavy-duty,each of which requires in its manufacture the use of two machines,A and B.A regular model requires the use of machine A for 2 hours and machine B for 3 hours.A heavy-duty model requires the use of machine A for 3 hours and machine B for 3 hours.Machine A can be used at most 18 hours a day and machine B can be used at most 21 hours a day.If the profits on the regular and heavy-duty models are $20 and $15,respectively,and United Blimpo Co.can sell all it produces,how many of each model should be produced per day in order to realize maximum profit? What is the maximum profit per day?
Question
Use the simplex method to maximize
Z = 30x + 50y
subject to
2x + y ≤ 16
x + 2y ≤ 11
x + 3y ≤ 15
x,y ≥ 0
Question
Use the simplex method to minimize
Z = - Use the simplex method to minimize Z = -   + 2   subject to   +   ≤ 4 5   +   ≥ -12 2   + 5   ≥ -14 3   - 2   ≤ 17<div style=padding-top: 35px>
+ 2 Use the simplex method to minimize Z = -   + 2   subject to   +   ≤ 4 5   +   ≥ -12 2   + 5   ≥ -14 3   - 2   ≤ 17<div style=padding-top: 35px>
subject to Use the simplex method to minimize Z = -   + 2   subject to   +   ≤ 4 5   +   ≥ -12 2   + 5   ≥ -14 3   - 2   ≤ 17<div style=padding-top: 35px>
+ Use the simplex method to minimize Z = -   + 2   subject to   +   ≤ 4 5   +   ≥ -12 2   + 5   ≥ -14 3   - 2   ≤ 17<div style=padding-top: 35px>
≤ 4
5 Use the simplex method to minimize Z = -   + 2   subject to   +   ≤ 4 5   +   ≥ -12 2   + 5   ≥ -14 3   - 2   ≤ 17<div style=padding-top: 35px>
+ Use the simplex method to minimize Z = -   + 2   subject to   +   ≤ 4 5   +   ≥ -12 2   + 5   ≥ -14 3   - 2   ≤ 17<div style=padding-top: 35px>
≥ -12
2 Use the simplex method to minimize Z = -   + 2   subject to   +   ≤ 4 5   +   ≥ -12 2   + 5   ≥ -14 3   - 2   ≤ 17<div style=padding-top: 35px>
+ 5 Use the simplex method to minimize Z = -   + 2   subject to   +   ≤ 4 5   +   ≥ -12 2   + 5   ≥ -14 3   - 2   ≤ 17<div style=padding-top: 35px>
≥ -14
3 Use the simplex method to minimize Z = -   + 2   subject to   +   ≤ 4 5   +   ≥ -12 2   + 5   ≥ -14 3   - 2   ≤ 17<div style=padding-top: 35px>
- 2 Use the simplex method to minimize Z = -   + 2   subject to   +   ≤ 4 5   +   ≥ -12 2   + 5   ≥ -14 3   - 2   ≤ 17<div style=padding-top: 35px>
≤ 17
Question
Use the dual and the simplex method to minimize
Z = 4 Use the dual and the simplex method to minimize Z = 4   + 5   subject to   -   ≥ 4 2   -   ≥ 1 5   + 3   ≥ 3   ,   ≥ 0.<div style=padding-top: 35px>
+ 5 Use the dual and the simplex method to minimize Z = 4   + 5   subject to   -   ≥ 4 2   -   ≥ 1 5   + 3   ≥ 3   ,   ≥ 0.<div style=padding-top: 35px>
subject to Use the dual and the simplex method to minimize Z = 4   + 5   subject to   -   ≥ 4 2   -   ≥ 1 5   + 3   ≥ 3   ,   ≥ 0.<div style=padding-top: 35px>
- Use the dual and the simplex method to minimize Z = 4   + 5   subject to   -   ≥ 4 2   -   ≥ 1 5   + 3   ≥ 3   ,   ≥ 0.<div style=padding-top: 35px>
≥ 4
2 Use the dual and the simplex method to minimize Z = 4   + 5   subject to   -   ≥ 4 2   -   ≥ 1 5   + 3   ≥ 3   ,   ≥ 0.<div style=padding-top: 35px>
- Use the dual and the simplex method to minimize Z = 4   + 5   subject to   -   ≥ 4 2   -   ≥ 1 5   + 3   ≥ 3   ,   ≥ 0.<div style=padding-top: 35px>
≥ 1
5 Use the dual and the simplex method to minimize Z = 4   + 5   subject to   -   ≥ 4 2   -   ≥ 1 5   + 3   ≥ 3   ,   ≥ 0.<div style=padding-top: 35px>
+ 3 Use the dual and the simplex method to minimize Z = 4   + 5   subject to   -   ≥ 4 2   -   ≥ 1 5   + 3   ≥ 3   ,   ≥ 0.<div style=padding-top: 35px>
≥ 3 Use the dual and the simplex method to minimize Z = 4   + 5   subject to   -   ≥ 4 2   -   ≥ 1 5   + 3   ≥ 3   ,   ≥ 0.<div style=padding-top: 35px>
, Use the dual and the simplex method to minimize Z = 4   + 5   subject to   -   ≥ 4 2   -   ≥ 1 5   + 3   ≥ 3   ,   ≥ 0.<div style=padding-top: 35px>
≥ 0.
Question
Find the dual problem to the following: A company has two different locations to assemble three different models of PCs.The table below summarizes the daily production capacity,the minimum number of each type needed,and the daily operating costs for each location.What is the number of days that each location needs to operate in order to fill the orders at minimum cost. Find the dual problem to the following: A company has two different locations to assemble three different models of PCs.The table below summarizes the daily production capacity,the minimum number of each type needed,and the daily operating costs for each location.What is the number of days that each location needs to operate in order to fill the orders at minimum cost.  <div style=padding-top: 35px>
Question
The dual of
Minimize
Z = <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. <div style=padding-top: 35px>
+ 3 <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. <div style=padding-top: 35px>
Subject to <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. <div style=padding-top: 35px>
- 2 <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. <div style=padding-top: 35px>
≥ 4
3 <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. <div style=padding-top: 35px>
+ <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. <div style=padding-top: 35px>
≥ 1 <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. <div style=padding-top: 35px>
, <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. <div style=padding-top: 35px>
≥ 0
Is:

A) Maximize W = <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. <div style=padding-top: 35px> + 3 <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. <div style=padding-top: 35px> subject to <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. <div style=padding-top: 35px> - 2 <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. <div style=padding-top: 35px> ≤ 4; 3 <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. <div style=padding-top: 35px> + <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. <div style=padding-top: 35px> ≤ 1; <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. <div style=padding-top: 35px> , <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. <div style=padding-top: 35px> ≥ 0.
B) Maximize W = <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. <div style=padding-top: 35px> + 3 <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. <div style=padding-top: 35px> subject to <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. <div style=padding-top: 35px> + 3 <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. <div style=padding-top: 35px> ≥ 4; - 2 <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. <div style=padding-top: 35px> + <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. <div style=padding-top: 35px> ≥ 1; <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. <div style=padding-top: 35px> , <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. <div style=padding-top: 35px> ≥ 0.
C) Maximize W = 4 <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. <div style=padding-top: 35px> + <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. <div style=padding-top: 35px> subject to <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. <div style=padding-top: 35px> - 2 <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. <div style=padding-top: 35px> ≥ 1; 3 <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. <div style=padding-top: 35px> + <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. <div style=padding-top: 35px> ≥ 3; <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. <div style=padding-top: 35px> , <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. <div style=padding-top: 35px> , ≥ 0.
D) Maximize W = 4 <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. <div style=padding-top: 35px> + <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. <div style=padding-top: 35px> subject to <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. <div style=padding-top: 35px> + 3 <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. <div style=padding-top: 35px> ≤ 1; - 2 <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. <div style=padding-top: 35px> + <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. <div style=padding-top: 35px> ≤ 3; <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. <div style=padding-top: 35px> , <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. <div style=padding-top: 35px> ≥ 0.
E) Maximize W = 4 <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. <div style=padding-top: 35px> + <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. <div style=padding-top: 35px> subject to <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. <div style=padding-top: 35px> + 3 <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. <div style=padding-top: 35px> ≥ 1; - 2 <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. <div style=padding-top: 35px> + <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. <div style=padding-top: 35px> ≥ 3; <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. <div style=padding-top: 35px> , <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. <div style=padding-top: 35px> ≥ 0.
Question
Find the dual problem to the following: The What If Company has $30,000 for the purchase of material to make three types of gadgets.The company has allocated a total of 1200 hours of assembly time and 180 hours of packaging time for the gadgets.The following table gives the cost per gadget,the number of hours per gadget,and the profit per gadget for each type. Find the dual problem to the following: The What If Company has $30,000 for the purchase of material to make three types of gadgets.The company has allocated a total of 1200 hours of assembly time and 180 hours of packaging time for the gadgets.The following table gives the cost per gadget,the number of hours per gadget,and the profit per gadget for each type.   What is the number of gadgets of each type the company should produce to maximize profit?<div style=padding-top: 35px>
What is the number of gadgets of each type the company should produce to maximize profit?
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Deck 8: Linear Programming
1
Sketch the region described by the inequalities. Sketch the region described by the inequalities.
Sketch the region described by the inequalities.
-3x + y = -3 -3x + y = -3      -3x + y = -3      -3x + y = -3
2
x - y > -3 <strong>x - y > -3  </strong> A)   B)   C)   D)

A) <strong>x - y > -3  </strong> A)   B)   C)   D)
B) <strong>x - y > -3  </strong> A)   B)   C)   D)
C) <strong>x - y > -3  </strong> A)   B)   C)   D)
D) <strong>x - y > -3  </strong> A)   B)   C)   D)
3
Maximize
Z = 5x - 3y
subject to
2x - y ≤ 8
2x - 5y ≥ 0
x - y = -2
x,y ≥ 0.
no optimum solution (empty feasible region)
4
To make some extra money,you make two types of picture frames,type A and type B,for sale.You have an initial start-up expense of $75.The production cost for type A is $3.60 per frame,and the production cost for type B is $5.20 per frame.The price for type A is $6.00 per frame and the price for type B is $10.00 per frame.Let x be the number of type A and y be the number of type B produced and sold.Write an inequality describing revenue less than cost.Solve the inequality and describe the region.Also,describe what this means in terms of frames.
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5
A car rental company has $540,000 to purchase up to 25 new cars of two different models.One model costs $18,000 each and the other model costs $24,000 each.Write a system of linear inequalities to describe the situation.Let x represent the first model and y represent the second.Find the region described by the system of linear inequalities.
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6
Sketch the region described by the inequalities. Sketch the region described by the inequalities.
Sketch the region described by the inequalities.
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7
4x + y > -2 <strong>4x + y > -2  </strong> A)   B)   C)   D)

A) <strong>4x + y > -2  </strong> A)   B)   C)   D)
B) <strong>4x + y > -2  </strong> A)   B)   C)   D)
C) <strong>4x + y > -2  </strong> A)   B)   C)   D)
D) <strong>4x + y > -2  </strong> A)   B)   C)   D)
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8
The XYZ Corporation produces two models of home computers,the Alpha model and the Beta model.Let x be the number of Alpha models and y the number of Beta models produced at the San Antonio factory per week.If the factory can produce at the most 100 Beta models in a week,write an inequality to describe this situation.Describe the region for the inequality.What additional inequalities can you add to the situation?
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9
The region indicated in the diagram <strong>The region indicated in the diagram   is described by</strong> A)   B)   C)   D)   E)   is described by

A) <strong>The region indicated in the diagram   is described by</strong> A)   B)   C)   D)   E)
B) <strong>The region indicated in the diagram   is described by</strong> A)   B)   C)   D)   E)
C) <strong>The region indicated in the diagram   is described by</strong> A)   B)   C)   D)   E)
D) <strong>The region indicated in the diagram   is described by</strong> A)   B)   C)   D)   E)
E) <strong>The region indicated in the diagram   is described by</strong> A)   B)   C)   D)   E)
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10
Sketch the region described by the inequalities Sketch the region described by the inequalities
Sketch the region described by the inequalities
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11
4x - 2y ≤ 4 <strong>4x - 2y ≤ 4  </strong> A)   B)   C)   D)

A) <strong>4x - 2y ≤ 4  </strong> A)   B)   C)   D)
B) <strong>4x - 2y ≤ 4  </strong> A)   B)   C)   D)
C) <strong>4x - 2y ≤ 4  </strong> A)   B)   C)   D)
D) <strong>4x - 2y ≤ 4  </strong> A)   B)   C)   D)
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12
Maximize
Z = 4x + y
subject to
-x + y ≤ 2
3x + y ≤ 18
x,y ≥ 0.
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13
A chair company produces two models of chairs,the Sequoia and the Saratoga.The Sequoia model takes 3 hours to assemble and A chair company produces two models of chairs,the Sequoia and the Saratoga.The Sequoia model takes 3 hours to assemble and   hour to paint.The Saratoga model takes 2 hours to assemble and 1 hour to paint.The maximum number of hours available to assemble is 24 per day and the maximum number of hours available to paint is 8 per day. (a)If the company earns a profit of $20 per Sequoia model and $30 per Saratoga model,find the number of models produced per day in order to maximize profit. (b)If the company earns a profit of $30 per Sequoia model and $15 per Saratoga model,find the number of models produced per day in order to maximize profit. (c)Suppose the company decides to upgrade the two models so it takes an additional 2 hours to detail the Sequoia and 2 hours to detail the Saratoga.The maximum number of hours available to detail is 18 per day.If the company earns a profit of $45 per Sequoia model and $35 per Saratoga model,find the number of models produced per day in order to maximize profit. (d)Suppose the company decides to upgrade the two models so it takes an additional 2 hours to detail the Sequoia and 2 hours to detail the Saratoga.The maximum number of hours available to detail is 18 per day.If the company earns a profit of $30 per Sequoia model and $40 per Saratoga model,find the number of models produced per day in order to maximize profit.
hour to paint.The Saratoga model takes 2 hours to assemble and 1 hour to paint.The maximum number of hours available to assemble is 24 per day and the maximum number of hours available to paint is 8 per day.
(a)If the company earns a profit of $20 per Sequoia model and $30 per Saratoga model,find the number of models produced per day in order to maximize profit.
(b)If the company earns a profit of $30 per Sequoia model and $15 per Saratoga model,find the number of models produced per day in order to maximize profit.
(c)Suppose the company decides to upgrade the two models so it takes an additional 2 hours to detail the Sequoia and 2 hours to detail the Saratoga.The maximum number of hours available to detail is 18 per day.If the company earns a profit of $45 per Sequoia model and $35 per Saratoga model,find the number of models produced per day in order to maximize profit.
(d)Suppose the company decides to upgrade the two models so it takes an additional 2 hours to detail the Sequoia and 2 hours to detail the Saratoga.The maximum number of hours available to detail is 18 per day.If the company earns a profit of $30 per Sequoia model and $40 per Saratoga model,find the number of models produced per day in order to maximize profit.
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14
The region indicated in the diagram <strong>The region indicated in the diagram   is described by</strong> A)   B)   C)   D)   E)   is described by

A) <strong>The region indicated in the diagram   is described by</strong> A)   B)   C)   D)   E)
B) <strong>The region indicated in the diagram   is described by</strong> A)   B)   C)   D)   E)
C) <strong>The region indicated in the diagram   is described by</strong> A)   B)   C)   D)   E)
D) <strong>The region indicated in the diagram   is described by</strong> A)   B)   C)   D)   E)
E) <strong>The region indicated in the diagram   is described by</strong> A)   B)   C)   D)   E)
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15
Maximize
Z = 2x - 3y
subject to
2x + y ≥ 1
x - y ≤ 1
x,y ≥ 0.
Also find the corner point where the value of z is attained.
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16
Sketch the region described by the following system of inequalities: Sketch the region described by the following system of inequalities:
Sketch the region described by the following system of inequalities:
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17
Sketch the region described by the inequalities. Sketch the region described by the inequalities.
Sketch the region described by the inequalities.
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18
A store sells two types of calculators.In order to cover overhead,it must sell at least 40 calculators total per week,and in order to satisfy distribution requirements,it must sell at least twice as many of type II as type I.Write a system of inequalities to describe the situation.Let x be the number of type I that it sells in a week and y be the number of type II that it sells in a week.Find the region described by the system of linear inequalities.
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19
2x - 3y ≥ 6 <strong>2x - 3y ≥ 6  </strong> A)   B)   C)   D)

A) <strong>2x - 3y ≥ 6  </strong> A)   B)   C)   D)
B) <strong>2x - 3y ≥ 6  </strong> A)   B)   C)   D)
C) <strong>2x - 3y ≥ 6  </strong> A)   B)   C)   D)
D) <strong>2x - 3y ≥ 6  </strong> A)   B)   C)   D)
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20
x + y ≤ -2 <strong>x + y ≤ -2  </strong> A)   B)   C)   D)

A) <strong>x + y ≤ -2  </strong> A)   B)   C)   D)
B) <strong>x + y ≤ -2  </strong> A)   B)   C)   D)
C) <strong>x + y ≤ -2  </strong> A)   B)   C)   D)
D) <strong>x + y ≤ -2  </strong> A)   B)   C)   D)
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21
Maximize
Z = <strong>Maximize Z =   - 2   + 3   Subject to  </strong> A) 0 B) 5 C) 10 D) 15 E) 20
- 2 <strong>Maximize Z =   - 2   + 3   Subject to  </strong> A) 0 B) 5 C) 10 D) 15 E) 20
+ 3 <strong>Maximize Z =   - 2   + 3   Subject to  </strong> A) 0 B) 5 C) 10 D) 15 E) 20
Subject to <strong>Maximize Z =   - 2   + 3   Subject to  </strong> A) 0 B) 5 C) 10 D) 15 E) 20

A) 0
B) 5
C) 10
D) 15
E) 20
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22
When no quotients exist in a simplex table,

A) a degenerate BFS will occur.
B) the problem has an unbounded solution.
C) there is a possibility for multiple optimum solutions.
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23
Use the simplex method to minimize
Z = 3 <strong>Use the simplex method to minimize Z = 3   + 2   Subject to       ,   ≥ 0</strong> A) Z =   when   =   ,   = 0 B) Z = 21 when   = 7,   = 0 C) Z = 14 when   = 0,   = 7 D) Z = 14 when   = 7,   = 0
+ 2 <strong>Use the simplex method to minimize Z = 3   + 2   Subject to       ,   ≥ 0</strong> A) Z =   when   =   ,   = 0 B) Z = 21 when   = 7,   = 0 C) Z = 14 when   = 0,   = 7 D) Z = 14 when   = 7,   = 0
Subject to <strong>Use the simplex method to minimize Z = 3   + 2   Subject to       ,   ≥ 0</strong> A) Z =   when   =   ,   = 0 B) Z = 21 when   = 7,   = 0 C) Z = 14 when   = 0,   = 7 D) Z = 14 when   = 7,   = 0
<strong>Use the simplex method to minimize Z = 3   + 2   Subject to       ,   ≥ 0</strong> A) Z =   when   =   ,   = 0 B) Z = 21 when   = 7,   = 0 C) Z = 14 when   = 0,   = 7 D) Z = 14 when   = 7,   = 0
<strong>Use the simplex method to minimize Z = 3   + 2   Subject to       ,   ≥ 0</strong> A) Z =   when   =   ,   = 0 B) Z = 21 when   = 7,   = 0 C) Z = 14 when   = 0,   = 7 D) Z = 14 when   = 7,   = 0
, <strong>Use the simplex method to minimize Z = 3   + 2   Subject to       ,   ≥ 0</strong> A) Z =   when   =   ,   = 0 B) Z = 21 when   = 7,   = 0 C) Z = 14 when   = 0,   = 7 D) Z = 14 when   = 7,   = 0
≥ 0

A) Z = <strong>Use the simplex method to minimize Z = 3   + 2   Subject to       ,   ≥ 0</strong> A) Z =   when   =   ,   = 0 B) Z = 21 when   = 7,   = 0 C) Z = 14 when   = 0,   = 7 D) Z = 14 when   = 7,   = 0 when <strong>Use the simplex method to minimize Z = 3   + 2   Subject to       ,   ≥ 0</strong> A) Z =   when   =   ,   = 0 B) Z = 21 when   = 7,   = 0 C) Z = 14 when   = 0,   = 7 D) Z = 14 when   = 7,   = 0 = <strong>Use the simplex method to minimize Z = 3   + 2   Subject to       ,   ≥ 0</strong> A) Z =   when   =   ,   = 0 B) Z = 21 when   = 7,   = 0 C) Z = 14 when   = 0,   = 7 D) Z = 14 when   = 7,   = 0 , <strong>Use the simplex method to minimize Z = 3   + 2   Subject to       ,   ≥ 0</strong> A) Z =   when   =   ,   = 0 B) Z = 21 when   = 7,   = 0 C) Z = 14 when   = 0,   = 7 D) Z = 14 when   = 7,   = 0 = 0
B) Z = 21 when <strong>Use the simplex method to minimize Z = 3   + 2   Subject to       ,   ≥ 0</strong> A) Z =   when   =   ,   = 0 B) Z = 21 when   = 7,   = 0 C) Z = 14 when   = 0,   = 7 D) Z = 14 when   = 7,   = 0 = 7, <strong>Use the simplex method to minimize Z = 3   + 2   Subject to       ,   ≥ 0</strong> A) Z =   when   =   ,   = 0 B) Z = 21 when   = 7,   = 0 C) Z = 14 when   = 0,   = 7 D) Z = 14 when   = 7,   = 0 = 0
C) Z = 14 when <strong>Use the simplex method to minimize Z = 3   + 2   Subject to       ,   ≥ 0</strong> A) Z =   when   =   ,   = 0 B) Z = 21 when   = 7,   = 0 C) Z = 14 when   = 0,   = 7 D) Z = 14 when   = 7,   = 0 = 0, <strong>Use the simplex method to minimize Z = 3   + 2   Subject to       ,   ≥ 0</strong> A) Z =   when   =   ,   = 0 B) Z = 21 when   = 7,   = 0 C) Z = 14 when   = 0,   = 7 D) Z = 14 when   = 7,   = 0 = 7
D) Z = 14 when <strong>Use the simplex method to minimize Z = 3   + 2   Subject to       ,   ≥ 0</strong> A) Z =   when   =   ,   = 0 B) Z = 21 when   = 7,   = 0 C) Z = 14 when   = 0,   = 7 D) Z = 14 when   = 7,   = 0 = 7, <strong>Use the simplex method to minimize Z = 3   + 2   Subject to       ,   ≥ 0</strong> A) Z =   when   =   ,   = 0 B) Z = 21 when   = 7,   = 0 C) Z = 14 when   = 0,   = 7 D) Z = 14 when   = 7,   = 0 = 0
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24
Maximize
Z = 10x + 15y
subject to Maximize Z = 10x + 15y subject to
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25
A manufacturer produces two products,product A and product B.Both products require processing on Machines I and II.The number of hours needed to produce one unit is given by the following chart: A manufacturer produces two products,product A and product B.Both products require processing on Machines I and II.The number of hours needed to produce one unit is given by the following chart:   Machine I is available for at most 1150 hours and Machine II is available for at most 1100 hours.If the profit made on product A is $15 / unit and the profit made on product B is $30 / unit.Find the production level that will maximize profit and find the Maximum profit.
Machine I is available for at most 1150 hours and Machine II is available for at most 1100 hours.If the profit made on product A is $15 / unit and the profit made on product B is $30 / unit.Find the production level that will maximize profit and find the Maximum profit.
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26
A person decides to take two different dietary supplements.Each supplement contains two essential ingredients,A and B,for which there are minimum daily requirements,and each contains a third ingredient,C,which needs to be minimized.Find the amount of each supplement that the person should take each day in order to satisfy the requirements for A and B while minimizing C.Also find the amount of C the person takes each day. A person decides to take two different dietary supplements.Each supplement contains two essential ingredients,A and B,for which there are minimum daily requirements,and each contains a third ingredient,C,which needs to be minimized.Find the amount of each supplement that the person should take each day in order to satisfy the requirements for A and B while minimizing C.Also find the amount of C the person takes each day.
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27
Use the simplex method to
maximize
Z = Use the simplex method to maximize Z =   + 4   +   subject to   .
+ 4 Use the simplex method to maximize Z =   + 4   +   subject to   .
+ Use the simplex method to maximize Z =   + 4   +   subject to   .
subject to Use the simplex method to maximize Z =   + 4   +   subject to   .
.
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28
A summer camp wants to hire counselors and aides to fill its staffing needs at minimum cost.The average monthly salary of a counselor is $2400 and the average monthly salary of an aide is $1100.The camp can accommodate up to 45 staff members and needs at least 30 to run properly.They must have at least 10 aides,and may have up to 3 aides for every 2 counselors.How many counselors and how many aides should the camp hire to minimize cost?

A) 12 counselors and 18 aides
B) 27 counselors and 18 aides
C) 35 counselors and 10 aides
D) 18 counselors and 12 aides
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29
A company has two different locations to assemble three different models of PCs.The table below summarizes the daily production capacity,the minimum number of each type needed,and the daily operating costs for each location.Find the number of days that each location needs to operate in order to fill the orders at minimum cost. A company has two different locations to assemble three different models of PCs.The table below summarizes the daily production capacity,the minimum number of each type needed,and the daily operating costs for each location.Find the number of days that each location needs to operate in order to fill the orders at minimum cost.
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30
Give the dual of:
Maximize
Z = Give the dual of: Maximize Z =   + 3   + 4   subject to 3   +   -   ≤ 4 4   -   + 2   ≥ 1   ,   ,   ≥ 0.
+ 3 Give the dual of: Maximize Z =   + 3   + 4   subject to 3   +   -   ≤ 4 4   -   + 2   ≥ 1   ,   ,   ≥ 0.
+ 4 Give the dual of: Maximize Z =   + 3   + 4   subject to 3   +   -   ≤ 4 4   -   + 2   ≥ 1   ,   ,   ≥ 0.
subject to
3 Give the dual of: Maximize Z =   + 3   + 4   subject to 3   +   -   ≤ 4 4   -   + 2   ≥ 1   ,   ,   ≥ 0.
+ Give the dual of: Maximize Z =   + 3   + 4   subject to 3   +   -   ≤ 4 4   -   + 2   ≥ 1   ,   ,   ≥ 0.
- Give the dual of: Maximize Z =   + 3   + 4   subject to 3   +   -   ≤ 4 4   -   + 2   ≥ 1   ,   ,   ≥ 0.
≤ 4
4 Give the dual of: Maximize Z =   + 3   + 4   subject to 3   +   -   ≤ 4 4   -   + 2   ≥ 1   ,   ,   ≥ 0.
- Give the dual of: Maximize Z =   + 3   + 4   subject to 3   +   -   ≤ 4 4   -   + 2   ≥ 1   ,   ,   ≥ 0.
+ 2 Give the dual of: Maximize Z =   + 3   + 4   subject to 3   +   -   ≤ 4 4   -   + 2   ≥ 1   ,   ,   ≥ 0.
≥ 1 Give the dual of: Maximize Z =   + 3   + 4   subject to 3   +   -   ≤ 4 4   -   + 2   ≥ 1   ,   ,   ≥ 0.
, Give the dual of: Maximize Z =   + 3   + 4   subject to 3   +   -   ≤ 4 4   -   + 2   ≥ 1   ,   ,   ≥ 0.
, Give the dual of: Maximize Z =   + 3   + 4   subject to 3   +   -   ≤ 4 4   -   + 2   ≥ 1   ,   ,   ≥ 0.
≥ 0.
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31
In the initial simplex table below,find the pivot entry. <strong>In the initial simplex table below,find the pivot entry.      </strong> A) -1 B) 0 C) 10 D) 2 E) 6 <strong>In the initial simplex table below,find the pivot entry.      </strong> A) -1 B) 0 C) 10 D) 2 E) 6
<strong>In the initial simplex table below,find the pivot entry.      </strong> A) -1 B) 0 C) 10 D) 2 E) 6

A) -1
B) 0
C) 10
D) 2
E) 6
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32
In the initial simplex table below,find the departing variable. <strong>In the initial simplex table below,find the departing variable.      </strong> A)   B)   C)   D)   E) Z <strong>In the initial simplex table below,find the departing variable.      </strong> A)   B)   C)   D)   E) Z
<strong>In the initial simplex table below,find the departing variable.      </strong> A)   B)   C)   D)   E) Z

A) <strong>In the initial simplex table below,find the departing variable.      </strong> A)   B)   C)   D)   E) Z
B) <strong>In the initial simplex table below,find the departing variable.      </strong> A)   B)   C)   D)   E) Z
C) <strong>In the initial simplex table below,find the departing variable.      </strong> A)   B)   C)   D)   E) Z
D) <strong>In the initial simplex table below,find the departing variable.      </strong> A)   B)   C)   D)   E) Z
E) Z
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33
A company manufactures two models of inline skates,Alpha and Beta,at two different manufacturing plants.The maximum output at plant I is 1000 per month,while the maximum output at plant II is 1200 per month.Due to contractual obligations,the number of Alpha models produced at plant I must exceed the number of Beta models produced at plant I by at least 100.The profit per Alpha and Beta model manufactured at plant I is $50 and $80,respectively,while the profit per Alpha and Beta model manufactured at plant II is $60 and $70,respectively.This month,the company received an order for 900 Alpha and 1000 Beta models.Find how many of each model should be produced at each plant in order to satisfy the order and maximize the profit.(Hint: Let A company manufactures two models of inline skates,Alpha and Beta,at two different manufacturing plants.The maximum output at plant I is 1000 per month,while the maximum output at plant II is 1200 per month.Due to contractual obligations,the number of Alpha models produced at plant I must exceed the number of Beta models produced at plant I by at least 100.The profit per Alpha and Beta model manufactured at plant I is $50 and $80,respectively,while the profit per Alpha and Beta model manufactured at plant II is $60 and $70,respectively.This month,the company received an order for 900 Alpha and 1000 Beta models.Find how many of each model should be produced at each plant in order to satisfy the order and maximize the profit.(Hint: Let   represent the number of Alpha models and   represent the number of Beta models manufactured at plant I.)
represent the number of Alpha models and A company manufactures two models of inline skates,Alpha and Beta,at two different manufacturing plants.The maximum output at plant I is 1000 per month,while the maximum output at plant II is 1200 per month.Due to contractual obligations,the number of Alpha models produced at plant I must exceed the number of Beta models produced at plant I by at least 100.The profit per Alpha and Beta model manufactured at plant I is $50 and $80,respectively,while the profit per Alpha and Beta model manufactured at plant II is $60 and $70,respectively.This month,the company received an order for 900 Alpha and 1000 Beta models.Find how many of each model should be produced at each plant in order to satisfy the order and maximize the profit.(Hint: Let   represent the number of Alpha models and   represent the number of Beta models manufactured at plant I.)
represent the number of Beta models manufactured at plant I.)
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34
A manufacturer produces two products,product A and product B.Both products require processing on Machines I and II.The number of hours needed to produce one unit is given by the following chart: A manufacturer produces two products,product A and product B.Both products require processing on Machines I and II.The number of hours needed to produce one unit is given by the following chart:   Machine I is available for at most 1000 hours and Machine II is available for at most 2500 hours.If the profit made on product A is $20 / unit and the profit made on product B is $25 / unit.Find the production level that will maximize profit and find the Maximum profit.
Machine I is available for at most 1000 hours and Machine II is available for at most 2500 hours.If the profit made on product A is $20 / unit and the profit made on product B is $25 / unit.Find the production level that will maximize profit and find the Maximum profit.
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35
Use the simplex method to minimize
Z = 4 Use the simplex method to minimize Z = 4   +   subject to   +   ≥ 5   + 2   ≥ 8   ,   ≥ 0.
+ Use the simplex method to minimize Z = 4   +   subject to   +   ≥ 5   + 2   ≥ 8   ,   ≥ 0.
subject to Use the simplex method to minimize Z = 4   +   subject to   +   ≥ 5   + 2   ≥ 8   ,   ≥ 0.
+ Use the simplex method to minimize Z = 4   +   subject to   +   ≥ 5   + 2   ≥ 8   ,   ≥ 0.
≥ 5 Use the simplex method to minimize Z = 4   +   subject to   +   ≥ 5   + 2   ≥ 8   ,   ≥ 0.
+ 2 Use the simplex method to minimize Z = 4   +   subject to   +   ≥ 5   + 2   ≥ 8   ,   ≥ 0.
≥ 8 Use the simplex method to minimize Z = 4   +   subject to   +   ≥ 5   + 2   ≥ 8   ,   ≥ 0.
, Use the simplex method to minimize Z = 4   +   subject to   +   ≥ 5   + 2   ≥ 8   ,   ≥ 0.
≥ 0.
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36
Maximize
Z = 4x + 6y
subject to Maximize Z = 4x + 6y subject to
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37
Use the simplex method to
maximize
Z = 2 Use the simplex method to maximize Z = 2   - 3   subject to   +   ≥ 5   + 2   ≤ 8   ,   ≥ 0
- 3 Use the simplex method to maximize Z = 2   - 3   subject to   +   ≥ 5   + 2   ≤ 8   ,   ≥ 0
subject to Use the simplex method to maximize Z = 2   - 3   subject to   +   ≥ 5   + 2   ≤ 8   ,   ≥ 0
+ Use the simplex method to maximize Z = 2   - 3   subject to   +   ≥ 5   + 2   ≤ 8   ,   ≥ 0
≥ 5 Use the simplex method to maximize Z = 2   - 3   subject to   +   ≥ 5   + 2   ≤ 8   ,   ≥ 0
+ 2 Use the simplex method to maximize Z = 2   - 3   subject to   +   ≥ 5   + 2   ≤ 8   ,   ≥ 0
≤ 8 Use the simplex method to maximize Z = 2   - 3   subject to   +   ≥ 5   + 2   ≤ 8   ,   ≥ 0
, Use the simplex method to maximize Z = 2   - 3   subject to   +   ≥ 5   + 2   ≤ 8   ,   ≥ 0
≥ 0
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38
Use the simplex method to solve the following problem: United Blimpo Co.produces two types of exercise devices,regular and heavy-duty,each of which requires in its manufacture the use of two machines,A and B.A regular model requires the use of machine A for 2 hours and machine B for 3 hours.A heavy-duty model requires the use of machine A for 3 hours and machine B for 3 hours.Machine A can be used at most 18 hours a day and machine B can be used at most 21 hours a day.If the profits on the regular and heavy-duty models are $20 and $15,respectively,and United Blimpo Co.can sell all it produces,how many of each model should be produced per day in order to realize maximum profit? What is the maximum profit per day?
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39
Use the simplex method to maximize
Z = 30x + 50y
subject to
2x + y ≤ 16
x + 2y ≤ 11
x + 3y ≤ 15
x,y ≥ 0
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40
Use the simplex method to minimize
Z = - Use the simplex method to minimize Z = -   + 2   subject to   +   ≤ 4 5   +   ≥ -12 2   + 5   ≥ -14 3   - 2   ≤ 17
+ 2 Use the simplex method to minimize Z = -   + 2   subject to   +   ≤ 4 5   +   ≥ -12 2   + 5   ≥ -14 3   - 2   ≤ 17
subject to Use the simplex method to minimize Z = -   + 2   subject to   +   ≤ 4 5   +   ≥ -12 2   + 5   ≥ -14 3   - 2   ≤ 17
+ Use the simplex method to minimize Z = -   + 2   subject to   +   ≤ 4 5   +   ≥ -12 2   + 5   ≥ -14 3   - 2   ≤ 17
≤ 4
5 Use the simplex method to minimize Z = -   + 2   subject to   +   ≤ 4 5   +   ≥ -12 2   + 5   ≥ -14 3   - 2   ≤ 17
+ Use the simplex method to minimize Z = -   + 2   subject to   +   ≤ 4 5   +   ≥ -12 2   + 5   ≥ -14 3   - 2   ≤ 17
≥ -12
2 Use the simplex method to minimize Z = -   + 2   subject to   +   ≤ 4 5   +   ≥ -12 2   + 5   ≥ -14 3   - 2   ≤ 17
+ 5 Use the simplex method to minimize Z = -   + 2   subject to   +   ≤ 4 5   +   ≥ -12 2   + 5   ≥ -14 3   - 2   ≤ 17
≥ -14
3 Use the simplex method to minimize Z = -   + 2   subject to   +   ≤ 4 5   +   ≥ -12 2   + 5   ≥ -14 3   - 2   ≤ 17
- 2 Use the simplex method to minimize Z = -   + 2   subject to   +   ≤ 4 5   +   ≥ -12 2   + 5   ≥ -14 3   - 2   ≤ 17
≤ 17
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41
Use the dual and the simplex method to minimize
Z = 4 Use the dual and the simplex method to minimize Z = 4   + 5   subject to   -   ≥ 4 2   -   ≥ 1 5   + 3   ≥ 3   ,   ≥ 0.
+ 5 Use the dual and the simplex method to minimize Z = 4   + 5   subject to   -   ≥ 4 2   -   ≥ 1 5   + 3   ≥ 3   ,   ≥ 0.
subject to Use the dual and the simplex method to minimize Z = 4   + 5   subject to   -   ≥ 4 2   -   ≥ 1 5   + 3   ≥ 3   ,   ≥ 0.
- Use the dual and the simplex method to minimize Z = 4   + 5   subject to   -   ≥ 4 2   -   ≥ 1 5   + 3   ≥ 3   ,   ≥ 0.
≥ 4
2 Use the dual and the simplex method to minimize Z = 4   + 5   subject to   -   ≥ 4 2   -   ≥ 1 5   + 3   ≥ 3   ,   ≥ 0.
- Use the dual and the simplex method to minimize Z = 4   + 5   subject to   -   ≥ 4 2   -   ≥ 1 5   + 3   ≥ 3   ,   ≥ 0.
≥ 1
5 Use the dual and the simplex method to minimize Z = 4   + 5   subject to   -   ≥ 4 2   -   ≥ 1 5   + 3   ≥ 3   ,   ≥ 0.
+ 3 Use the dual and the simplex method to minimize Z = 4   + 5   subject to   -   ≥ 4 2   -   ≥ 1 5   + 3   ≥ 3   ,   ≥ 0.
≥ 3 Use the dual and the simplex method to minimize Z = 4   + 5   subject to   -   ≥ 4 2   -   ≥ 1 5   + 3   ≥ 3   ,   ≥ 0.
, Use the dual and the simplex method to minimize Z = 4   + 5   subject to   -   ≥ 4 2   -   ≥ 1 5   + 3   ≥ 3   ,   ≥ 0.
≥ 0.
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42
Find the dual problem to the following: A company has two different locations to assemble three different models of PCs.The table below summarizes the daily production capacity,the minimum number of each type needed,and the daily operating costs for each location.What is the number of days that each location needs to operate in order to fill the orders at minimum cost. Find the dual problem to the following: A company has two different locations to assemble three different models of PCs.The table below summarizes the daily production capacity,the minimum number of each type needed,and the daily operating costs for each location.What is the number of days that each location needs to operate in order to fill the orders at minimum cost.
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43
The dual of
Minimize
Z = <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0.
+ 3 <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0.
Subject to <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0.
- 2 <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0.
≥ 4
3 <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0.
+ <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0.
≥ 1 <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0.
, <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0.
≥ 0
Is:

A) Maximize W = <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. + 3 <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. subject to <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. - 2 <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. ≤ 4; 3 <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. + <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. ≤ 1; <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. , <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. ≥ 0.
B) Maximize W = <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. + 3 <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. subject to <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. + 3 <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. ≥ 4; - 2 <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. + <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. ≥ 1; <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. , <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. ≥ 0.
C) Maximize W = 4 <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. + <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. subject to <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. - 2 <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. ≥ 1; 3 <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. + <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. ≥ 3; <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. , <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. , ≥ 0.
D) Maximize W = 4 <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. + <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. subject to <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. + 3 <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. ≤ 1; - 2 <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. + <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. ≤ 3; <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. , <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. ≥ 0.
E) Maximize W = 4 <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. + <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. subject to <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. + 3 <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. ≥ 1; - 2 <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. + <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. ≥ 3; <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. , <strong>The dual of Minimize Z =   + 3   Subject to   - 2   ≥ 4 3   +   ≥ 1   ,   ≥ 0 Is:</strong> A) Maximize W =   + 3   subject to   - 2   ≤ 4; 3   +   ≤ 1;   ,   ≥ 0. B) Maximize W =   + 3   subject to   + 3   ≥ 4; - 2   +   ≥ 1;   ,   ≥ 0. C) Maximize W = 4   +   subject to   - 2   ≥ 1; 3   +   ≥ 3;   ,   , ≥ 0. D) Maximize W = 4   +   subject to   + 3   ≤ 1; - 2   +   ≤ 3;   ,   ≥ 0. E) Maximize W = 4   +   subject to   + 3   ≥ 1; - 2   +   ≥ 3;   ,   ≥ 0. ≥ 0.
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Find the dual problem to the following: The What If Company has $30,000 for the purchase of material to make three types of gadgets.The company has allocated a total of 1200 hours of assembly time and 180 hours of packaging time for the gadgets.The following table gives the cost per gadget,the number of hours per gadget,and the profit per gadget for each type. Find the dual problem to the following: The What If Company has $30,000 for the purchase of material to make three types of gadgets.The company has allocated a total of 1200 hours of assembly time and 180 hours of packaging time for the gadgets.The following table gives the cost per gadget,the number of hours per gadget,and the profit per gadget for each type.   What is the number of gadgets of each type the company should produce to maximize profit?
What is the number of gadgets of each type the company should produce to maximize profit?
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