Chapter 7.4, Problem 35E

Solution

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

The minimum value is 24.

Given Information:

The system of inequality is,

  $\begin{array}{c}\text{Objective function}:f=5x+2y\\ \text{Constraints:}\left\{\begin{array}{c}2x+y\ge 12\\ 4x+3y\ge 30\\ x+2y\ge 10\end{array}\right.\\ x\ge 0,y\ge 0\end{array}$

Calculation:

Consider the given system of inequality,

  $\begin{array}{c}\text{Objective function}:f=5x+2y\\ \text{Constraints:}\left\{\begin{array}{c}2x+y\ge 12\\ 4x+3y\ge 30\\ x+2y\ge 10\end{array}\right.\\ x\ge 0,y\ge 0\end{array}$

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.

  $\begin{array}{c}2x+y=12\\ 4x+3y=30\\ x+2y=10\end{array}$

Find the intersecting points between the equations.

Substitute $x=10-2y$ in the second equation.

  $\begin{array}{c}4\left(10-2y\right)+3y=30\\ 40-8y+3y=30\\ 10=5y\\ y=2\end{array}$

The obtained point is $\left(6,2\right)$ .

Substitute $y=12-2x$ in the second equation.

  $\begin{array}{c}4x+3\left(12-2x\right)=30\\ 4x+36-6x=30\\ 6=2x\\ x=3\end{array}$

The obtained point is $\left(3,6\right)$ .

Substitute $x=0$ in the first equation.

  $\begin{array}{c}2\left(0\right)+y=12\\ y=12\end{array}$

The obtained point is $\left(0,12\right)$ .

Substitute $y=0$ in the third equation.

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

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

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 12 for y in the objective function.

  $\begin{array}{c}f=5\left(0\right)+2\left(12\right)\\ =24\end{array}$

Substitute 3 for x and 6 for y in the objective function.

  $\begin{array}{c}f=5\left(3\right)+2\left(6\right)\\ =27\end{array}$

Substitute 6 for x and 2 for y in the objective function.

  $\begin{array}{c}f=5\left(6\right)+2\left(2\right)\\ =34\end{array}$

Substitute 10 for x and 0 for y in the objective function.

  $\begin{array}{c}f=5\left(10\right)+2\left(0\right)\\ =50\end{array}$

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 24.