Chapter 7.6, Problem 20E

To determine

Todo:Graph the region corresponding to the system of constraints given below. Then find the minimum and maximum values of the objective function and the points where they occur, subject to the constraints.

Objective function:

  $z=y$

Constraints:

  $x\ge 0$

  $y\ge 0$

  $2x+3y\le 60$

  $2x+y\le 28$

  $4x+y\le 48$

Answer to Problem 20E

The region corresponding to the system of constraints is

  

The minimum value of $z$ is $0$ , at $\left(x,y\right)=\left(0,0\right)$ and $\left(x,y\right)=\left(12,0\right)$

The maximum value of $z$ is $20$ , at $\left(x,y\right)=\left(0,20\right)$

Explanation of Solution

Given:

Objective function:

  $z=y$

Constraints:

  $x\ge 0$

  $y\ge 0$

  $2x+3y\le 60$

  $2x+y\le 28$

  $4x+y\le 48$

Concept Used:

To find the minimum and maximum values of the objective function.

First plot the given constraints on the graph and find the bounded region. Then evaluating the objective function at each of the vertex of the bounded region results into the minimum and maximum value of the function.

Calculation:

For the given

Objective function:

  $z=y$

Constraints:

  $x\ge 0$

  $y\ge 0$

  $2x+3y\le 60$

  $2x+y\le 28$

  $4x+y\le 48$

Plotting the given constraints on a graph,

Let $x\ge 0------\left(1\right)$

  $y\ge 0------\left(2\right)$

  $2x+3y\le 60------\left(3\right)$

  $2x+y\le 28------\left(4\right)$

  $4x+y\le 48------\left(5\right)$

Consider the equation $\left(3\right)$ and $\left(4\right)$ , as

  $2x+3y=60------\left(6\right)$

and

  $2x+y=28------\left(7\right)$

and

  $4x+y=48------\left(8\right)$

Consider $2x+3y=60$

At $x=0$

  $\begin{array}{c}2\left(0\right)+3y=60\\ 3y=60\\ y=\frac{60}{3}\\ y=20\end{array}$

At $y=0$

  $\begin{array}{c}2x+3\left(0\right)=60\\ 2x=60\\ x=\frac{60}{2}\\ x=30\end{array}$

$x$	$0$	$30$
$y$	$20$	$0$

Consider $2x+y=28$

At $x=0$

  $\begin{array}{c}2\left(0\right)+y=28\\ y=28\end{array}$

At $y=0$

  $\begin{array}{c}2x+0=28\\ x=\frac{28}{2}\\ x=14\end{array}$

$x$	$0$	$14$
$y$	$28$	$0$

Consider $4x+y=48$

At $x=0$

  $\begin{array}{c}4\left(0\right)+y=48\\ y=48\end{array}$

At $y=0$

  $\begin{array}{c}4x+0=48\\ x=\frac{48}{4}\\ x=12\end{array}$

$x$	$0$	$12$
$y$	$48$	$0$

Plotting the region bounded by the given constraints, we have

  

Evaluating the objective function at each vertex

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

  $z=0$

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

  $z=0$

At $\left(10,8\right)$

  $z=8$

At $\left(6,16\right)$

  $z=16$

At $\left(0,20\right)$

  $z=20$

The minimum value of $z$ is $0$ , at $\left(x,y\right)=\left(0,0\right)$ and $\left(x,y\right)=\left(12,0\right)$

The maximum value of $z$ is $20$ , at $\left(x,y\right)=\left(0,20\right)$

Conclusion:

The region corresponding to the system of constraints is

  

The minimum value of $z$ is $0$ , at $\left(x,y\right)=\left(0,0\right)$ and $\left(x,y\right)=\left(12,0\right)$

The maximum value of $z$ is $20$ , at $\left(x,y\right)=\left(0,20\right)$