An objective function and a system of linear inequalities representing constraints are given. Graph the system of
inequalities representing the constraints. Find the value of the objective function at each corner of the graphed
region. Use these values to determine the maximum value of the objective function and the values of x and y for
which the maximum occurs.
-$$\begin{array} { l l }
\text { Objective Function } & z = 6 x + 7 y \
\text { Constraints } & x \geq 0 \
& y \geq 0 \
& 2 x + 3 y \leq 12 \
& 2 x + y \leq 8
\end{array}$$
A)Maximum 32; at (3, 2)
B)Maximum 32; at (2, 3)
C)Maximum 52; at (4, 4)
D)Maximum 24; at (4, 0)

Question

An objective function and a system of linear inequalities representing constraints are given. Graph the system of
inequalities representing the constraints. Find the value of the objective function at each corner of the graphed
region. Use these values to determine the maximum value of the objective function and the values of x and y for
which the maximum occurs.
-$$\begin{array} { l l } 
\text { Objective Function } & z = 6 x + 7 y \
\text { Constraints } & x \geq 0 \
& y \geq 0 \
& 2 x + 3 y \leq 12 \
& 2 x + y \leq 8
\end{array}$$
A)Maximum 32; at (3, 2)
B)Maximum 32; at (2, 3)
C)Maximum 52; at (4, 4)
D)Maximum 24; at (4, 0)

Quizplus · Accepted Answer

First, we graph the system of inequalities by finding the equations of the boundary lines:

$x=0$: Vertical line passing through the origin.

$y=0$: Horizontal line passing through the origin.

$2x+3y=12$: Solving for y, we get $y=-\frac{2}{3}x+4$. To graph this line, we can plot the intercepts $(0,4)$ and $(6,0)$ and connect them.

$2x+y=8$: Solving for y, we get $y=-2x+8$. To graph this line, we can plot the intercepts $(0,8)$ and $(4, 0)$ and connect them.

The graph of the system of inequalities looks like this:

\begin{array}{c|c|c|c} 
 & x\geq0 & y \geq 0 & 2x+3y \leq 12 \ 
 	ext{Region 1} & \large{	imes} & \large{	imes} & \blacksquare \  
 	ext{Region 2} & \blacksquare & \large{	imes} & \large{	imes} \ 
 	ext{Region 3} & \blacksquare & \blacksquare & \large{	imes} \ 
 	ext{Region 4} & \large{	imes} & \blacksquare & \large{	imes} \ 
 \end{array}

Next, we find the corner points of the shaded region. These are the points where the boundary lines intersect.

Corner point 1: (0, 4) where the lines $y=0$ and $2x+3y=12$ intersect.

Corner point 2: (2, 2) where the lines $2x+y=8$ and $2x+3y=12$ intersect.

Corner point 3: (4, 0) where the lines $x=0$ and $2x+y=8$ intersect.

Now we substitute each corner point into the objective function to find the maximum value:

Corner point 1: $z=6(0)+7(4)=28$

Corner point 2: $z=6(2)+7(2)=20+14=34$

Corner point 3: $z=6(4)+7(0)=24$

The maximum value of the objective function is 34, which occurs at corner point 2, (2, 2). Therefore, the answer is choice A.

An Objective Function and a System of Linear Inequalities Representing