Chapter 7.4, Problem 36E

Solution

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

The minimum value is 34.

Given Information:

The system of inequality is,

  $\begin{array}{c}\text{Objective function}:f=3x+5y\\ \text{Constraints:}\left\{\begin{array}{c}3x+2y\ge 20\\ 5x+6y\ge 52\\ 2x+7y\ge 30\end{array}\right.\\ x\ge 0,y\ge 0\end{array}$

Calculation:

Consider the given system of inequality,

  $\begin{array}{c}\text{Objective function}:f=3x+5y\\ \text{Constraints:}\left\{\begin{array}{c}3x+2y\ge 20\\ 5x+6y\ge 52\\ 2x+7y\ge 30\end{array}\right.\\ x\ge 0,y\ge 0\end{array}$

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.

  $\begin{array}{l}3x+2y=20\\ 5x+6y=52\\ 2x+7y=30\end{array}$

Find the intersecting points between the equations.

Substitute $x=0$ in the first equation.

  $\begin{array}{c}3\left(0\right)+2y=20\\ 2y=20\\ y=10\end{array}$

The obtained point is $\left(0,10\right)$ .

Substitute $y=\frac{20-3x}{2}$ in the second equation.

  $\begin{array}{c}5x+6\left(\frac{20-3x}{2}\right)=52\\ \frac{10x+120-18x}{2}=52\\ 120-104=8x\\ x=2\end{array}$

The obtained point is $\left(2,7\right)$ .

Substitute $x=\frac{30-7y}{2}$ in the second equation.

  $\begin{array}{c}5\left(\frac{30-7y}{2}\right)+6y=52\\ 150-35y+12y=104\\ 46=23y\\ y=2\end{array}$

The obtained point is $\left(8,2\right)$ .

Substitute $y=0$ in the third equation.

  $\begin{array}{c}2x+7\left(0\right)=30\\ 2x=30\\ x=15\end{array}$

The obtained point is $\left(15,0\right)$ .

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 10 for y in the objective function.

  $\begin{array}{c}f=3\left(0\right)+5\left(10\right)\\ =50\end{array}$

Substitute 2 for x and 7 for y in the objective function.

  $\begin{array}{c}f=3\left(2\right)+5\left(7\right)\\ =41\end{array}$

Substitute 8 for x and 2 for y in the objective function.

  $\begin{array}{c}f=3\left(8\right)+5\left(2\right)\\ =34\end{array}$

Substitute 15 for x and 0 for y in the objective function.

  $\begin{array}{c}f=3\left(15\right)+5\left(0\right)\\ =45\end{array}$

Since the feasible region is unbounded and has  $x\ge 0$  and  $y\ge 0$ , and conclude that there is no maximum value for the objective function.

Therefore, the minimum value is 34.