Chapter 7, Problem 18CT

To determine

To find: the minimum and maximum values of the objective function $z=20x+12y$ .

Answer to Problem 18CT

The minimum value of $z$is 0 and it occurs at when $x=0$ and $y=0$ .

The maximum value of $z$ is 240 and it occurs at when $x=12$and $y=0$ .

Explanation of Solution

Given information:

The given following constraints,

  $\begin{array}{c}x\ge 0\\ y\ge 0\\ x+4y\le 32\\ 3x+2y\le 36\end{array}$

Calculation:

Consider the objective function as

  $z=20x+12y$

Also, the above objective function subjects to the following constraints

  $\begin{array}{c}x\ge 0\\ y\ge 0\\ x+4y\le 32\\ 3x+2y\le 36\end{array}$

The inequalities $x\ge 0,y\ge 0$are sign constraints which implies that the feasible region lies in Quadrant I.

Consider the equations

  $\begin{array}{c}x+4y=32\\ 3x+2y=36\end{array}$

Plot these equations using Graphing Calculator

Step 1: Write the equations in the rectangular box as follows

  $x=0,y=0,x+4y=32,3x+2y=36$

Step 2: Right click on the equations, go to plot and click on 2 −D implict plot

The graphs of the equations $x=0,y=0,x+4y=32,3x+2y=36$ are shown below

  

The region bounded by the constraints is shown in the above plot.

Test objective function at each vertex

At $\left(0,0\right)$

  $\begin{array}{c}z=20\left(0\right)+12\left(0\right)\\ =0\end{array}$

At $\left(0,8\right)$

  $\begin{array}{c}z=20\left(0\right)+12\left(8\right)\\ =96\end{array}$

At $\left(12,0\right)$

  $\begin{array}{c}z=20\left(12\right)+12\left(0\right)\\ =240\end{array}$

At $\left(8,6\right)$

  $\begin{array}{c}z=20\left(8\right)+12\left(6\right)\\ =160+72\\ =232\end{array}$

Hence, the minimum value of $z$is 0 and it occurs at when $x=0$ and $y=0$ .

Also, the maximum value of $z$ is 240 and it occurs at when $x=12$and $y=0$ .