Chapter 3.3, Problem 22PPS

To determine

To find: The minimum value and the maximum value of the given function.

Answer to Problem 22PPS

The minimum value is $–18$ and the maximum value is $26$ .

Explanation of Solution

Given information :

The system of inequalities:

  $\begin{array}{r}\hfill y\ge |x+1|-2\\ \hfill 0\le y\le 6\\ \hfill -6\le x\le 2\\ \hfill x+3y\le 14\\ \hfill f(x,y)=5x+4y\end{array}$

Calculation :

The system of inequality is written as follows:

  $\begin{array}{r}\hfill y\ge |x+1|-2\\ \hfill 0\le y\le 6\\ \hfill -6\le x\le 2\\ \hfill x+3y\le 14\\ \hfill f(x,y)=5x+4y\end{array}$

The system of equation is plotted as shown below:

  

Region R is the feasible region, where all functions have solution. Therefore, the vertices of the feasible region are:

(−4,6),(2,4),(2,1),(1,0),(−3,0),(−6,3)and (−6,6)

Substitute the points $\left(–4,6\right),\left(2,4\right),\left(2,1\right),\left(1,0\right),\left(–3,0\right),\left(–6,3\right)$ and $\left(–6,6\right)$ in the function

  $f(x,y)=5x+4y$ as follows:

$\left(x,y\right)$	$f\left(x,y\right)=5x+4y$
(−4,6)	$f\left(–4,6\right)=5\left(–4\right)+4\left(6\right)=4$
$\left(2,4\right)$	$f\left(2,4\right)=5\left(2\right)+4\left(4\right)=26$
$\left(2,1\right)$	$f\left(2,1\right)=5\left(1\right)+4\left(1\right)=14$
$\left(1,0\right)$	$f\left(1,0\right)=5\left(1\right)+4\left(0\right)=5$
$\left(–3,0\right)$	$f\left(–3,0\right)=5\left(–3\right)+4\left(0\right)=–15$
$\left(–6,3\right)$	$f\left(–6,3\right)=5\left(–6\right)+4\left(3\right)=–18$
$\left(–6,6\right)$	$f\left(–6,6\right)=5\left(–6\right)+4\left(6\right)=6$

The minimum value is $–18$ and the maximum value is $26$ .