Chapter 7.4, Problem 31E

Solution

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

  $\begin{array}{c}\text{Objective function}:f=4x+3y\\ x+y\le 80\\ \text{Constraints:}x-2y\le 0\\ x\ge 0,y\ge 0\end{array}$

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

Given Information:

The system of inequality is,

  $\begin{array}{c}\text{Objective function}:f=4x+3y\\ x+y\le 80\\ \text{Constraints:}x-2y\le 0\\ x\ge 0,y\ge 0\end{array}$

Calculation:

Consider the given system of inequality,

  $\begin{array}{c}\text{Objective function}:f=4x+3y\\ x+y\le 80\\ \text{Constraints:}x-2y\le 0\\ x\ge 0,y\ge 0\end{array}$

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.

  $\begin{array}{c}x+y=80\\ x-2y=0\end{array}$

Substitute $2y$ for $x$ in the first equation.

  $\begin{array}{l}2y+y=80\\ 3y=80\\ y=26.67\end{array}$

And,

  $\begin{array}{c}x=2\left(26.67\right)\\ =53.34\end{array}$

The point is $\left(53.34,26.67\right)$ .

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.

  $\begin{array}{c}f=4\left(0\right)+3\left(0\right)\\ =0\end{array}$

Which is the minimum value.

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

  $\begin{array}{c}f=4\left(0\right)+3\left(80\right)\\ =240\end{array}$

Substitute $53.34$ for x and $26.67$ for y in the objective function.

  $\begin{array}{c}f=4\left(53.34\right)+3\left(26.67\right)\\ =\text{213}\text{.36}+\text{80}\text{.01}\\ =\text{293}\text{.37}\end{array}$

Which is the maximum value.

Hence, the minimum value is 0 and the maximum value is $293.37$ .