Chapter 7.4, Problem 32E

Solution

To calculate: The minimum and minimum value, checked they exists, of the objective function $f$ .

The common region in the graph is the solution of the inequality.

Given Information:

The system of inequality is,

  $\begin{array}{c}\text{Objective function}:f=10x+11y\\ x+y\le 90\\ \text{Constraints:3}x-y\ge 0\\ x\ge 0,y\ge 0\end{array}$

Calculation:

Consider the given system of inequality,

  $\begin{array}{c}\text{Objective function}:f=10x+11y\\ x+y\le 90\\ \text{Constraints:3}x-y\ge 0\\ x\ge 0,y\ge 0\end{array}$

Use the graphing calculator to draw the inequality.

  

The polygonal region of feasible solutions has three vertices. To find these, solve each of the following system of boundary equations:

Write the inequality as equation and find the intersecting point.

  $\begin{array}{c}x+y=90\\ 3x-y=0\end{array}$

Substitute $y$ for $3x$ in the first equation.

  $\begin{array}{c}x+3x=90\\ 4x=90\\ x=22.5\end{array}$

And,

  $\begin{array}{c}x=3\left(22.5\right)\\ =67.5\end{array}$

The point is $\left(22.5,67.5\right)$ .

So, mentioned the point on the graph and shaded the common region.

  

Find the value of the objective function at the all vertices.

Substitute 0 for x and 0 for y in the objective function.

  $\begin{array}{c}f=10\left(0\right)+11\left(0\right)\\ =0\end{array}$

Which is the minimum value.

Substitute 0 for x and 80 for y in the objective function.

  $\begin{array}{c}f=10\left(90\right)+11\left(0\right)\\ =900\end{array}$

Substitute $22.5$ for x and $67.5$ for y in the objective function.

  $\begin{array}{c}f=10\left(22.5\right)+11\left(67.5\right)\\ =\text{225}+\text{742}\text{.5}\\ =\text{967}\text{.5}\end{array}$

Which is the maximum value.

Hence, the minimum value is 0 and the maximum value is $\text{967}\text{.5}$ .