## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 7 - Review Exercises - Page 878: 56

#### Answer

The value of the objective function at the corner point $\left( 2,2 \right)$ is $10$, at the point $\left( 4,0 \right)$ is $8$, at the point $\left( \frac{1}{2},\frac{1}{2} \right)$ is $\frac{5}{2}$ and at the point $\left( 1,0 \right)$ is $2$. The maximum value of the objective function is $10$ at the point $\left( 2,2 \right)$ and the minimum value of the objective function is $2$ at the point $\left( 1,0 \right)$.

#### Work Step by Step

The corner points of the given graph are $\left( 2,2 \right)$, $\left( 4,0 \right)$, $\left( \frac{1}{2},\frac{1}{2} \right)$, $\left( 1,0 \right)$ and the objective function is $z=2x+3y$. The value of objective function at each corner point can be computed as given below. At the point $\left( 1,0 \right)$: \begin{align} & \left( 1,0 \right)=2\times 1+3\times 0 \\ & =2 \end{align} At the point $\left( \frac{1}{2},\frac{1}{2} \right)$: \begin{align} & z=2\times \frac{1}{2}+3\times \frac{1}{2} \\ & =\frac{5}{2} \end{align} Next, at the point $\left( 4,0 \right)$: \begin{align} & z=2\times 4+3\times 0 \\ & =8 \end{align} Also, at the point $\left( 2,2 \right)$: \begin{align} & z=2\times 2+3\times 2 \\ & =10 \end{align} Therefore, the maximum value of the objective function is $10$ at the point $\left( 2,2 \right)$, And, the minimum value of the objective function is $2$ at the point $\left( 1,0 \right)$. Thus, the value of the objective function at the point $\left( 2,2 \right)$ is $10$, at the point $\left( 4,0 \right)$ is $8$, at the point $\left( \frac{1}{2},\frac{1}{2} \right)$ is $\frac{5}{2}$, and at the point $\left( 1,0 \right)$ is $2$. The maximum value of the objective function is $10$ at the point $\left( 2,2 \right)$ and the minimum value of the objective function is $2$ at the point $\left( 1,0 \right)$.

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