Chapter 7.4, Problem 46E

Solution

The correction option for the solution of the system of inequality.

  $\begin{array}{l}\text{Objective function: }f=5x+10y\\ \text{Constraints:}\left\{\begin{array}{l}2x+y\le 10\\ x+3y\le 12\\ x\ge 0,y\ge 0\end{array}\right.\end{array}$

A. 0

B. 25

C. 40

D. 46

E. 55

The correct choice is D.

Given Information:

The system of inequality is defined as,

  $\begin{array}{l}\text{Objective function: }f=5x+10y\\ \text{Constraints:}\left\{\begin{array}{l}2x+y\le 10\\ x+3y\le 12\\ x\ge 0,y\ge 0\end{array}\right.\end{array}$

Explanation:

Consider the given inequality,

  $\begin{array}{l}\text{Objective function: }f=5x+10y\\ \text{Constraints:}\left\{\begin{array}{l}2x+y\le 10\\ x+3y\le 12\\ x\ge 0,y\ge 0\end{array}\right.\end{array}$

Draw the inequality by using the graphing calculator.

  

The find the maximum value in the feasible region, find the value of the objective function at the common boundary point.

Use one by one to check the value.

  $\begin{array}{c}f=5\left(0\right)+10\left(0\right)\left(0,0\right)\\ =0+0\\ =0\end{array}$

And,

  $\begin{array}{c}f=5\left(5\right)+10\left(0\right)\left(5,0\right)\\ =25+0\\ =25\end{array}$

And,

  $\begin{array}{c}f=5\left(0\right)+10\left(4\right)\left(0,4\right)\\ =0+40\\ =40\end{array}$

And,

  $\begin{array}{c}f=5\left(3.6\right)+10\left(2.8\right)\left(3.6,2.8\right)\\ =18.0+28\\ =46\end{array}$

The obtained maximum value is 46.

Hence, the required correct option is D.