Consider the following linear program: Maximize 30X + 10Y
Subject to: 3X + Y ≤ 300
X + Y ≤ 200
X ≤ 100
Y ≥ 50
X,Y ≥ 0
What is the optimal solution?
A)$3,000
B)$3,200
C)$3,400
D)$3,600

Question

Quizplus · Accepted Answer

To find the optimal solution, we need to solve the linear program. Using an online graphing calculator or software, we can plot the constraints on a graph and find the feasible region.
The feasible region is the shaded portion below:

[asy] size(8cm); import graph3; import solids; defaultpen(linewidth(0.7)); pen font = "Tahoma";  triple O=(0,0,0), X=(1,0,0), Y=(0,1,0), Z=(0,0,1), I=(1,1,0); draw(surface(((100,0)--(0,200)--(165,135)--cycle)),rgb(0.6,0.7,0.8),nolight); draw((100,0)--(0,200), linewidth(1.5)); draw((0,0)--(0,250),EndArrow); draw((0,0)--(250,0),EndArrow); label("$X$",(260,0,0),font=font); label("$Y$",(0,260,0),font=font); labelrotate(35,"$3X+Y=300$",(150,55,0),align=NW); labelrotate(138,"$X+Y=200$",(100,160,0),align=SW); label("$50$",(0,50,0), W, red); label("$100$",(100,0,0),S,red); label("$200$",(0,200,0),W,red); label("$O$",O,N,font=font); [/asy]

The vertices of the feasible region are (0,50), (100,0), and (65,135). We can evaluate the objective function 30X + 10Y at each vertex to determine which one gives the maximum value. 
(0,50): 30(0) + 10(50) = $500
(100,0): 30(100) + 10(0) = $3,000
(65,135): 30(65) + 10(135) = $3,400
Therefore, the optimal solution is when X = 100 and Y = 0, which gives a maximum value of $3,000. Thus, the answer is choice letter A.

Consider the Following Linear Program: Maximize 30X + 10Y