Chapter 3.10, Problem 4E

To determine

To Find: To find the maximum and minimum value of the given objective function.

Answer to Problem 4E

Maximum value is 36 and minimum value is $-38.$

Explanation of Solution

Given information: Given function $S=6x-2y$

Graph:

The given inequalities are $0\le y\le 10,y\ge -2x+4,\text{and}y\le -2x+12$

  

Calculation:

From the graph it can be seen that the feasible region is bounded by the points $(-3,10),(2,0),(1,10)\text{and}(6,0)$

Now, the given function is $S=6x-2y$

  $\begin{array}{l}\text{at}(-3,10),\text{we}\text{get}\\ S=6\times -3-2\times 10\\ S=-18-20\\ S=-38\\ \text{Thus,}\text{the}\text{value}\text{of}S\text{at}(-3,10)\text{is}-38.\\ \text{Similarly,}\\ \text{at}(2,0),\text{we}\text{get}\\ S=6\times 2-2\times 0\\ S=12-0\\ S=12\\ \therefore \text{at}(2,0)\text{the}\text{value}\text{of}S=12\\ \text{also,}\text{at}(1,10),\text{we}\text{get}\\ S=6\times 1-2\times 10\\ S=6-20\\ S=-14\\ \therefore \text{at}(1,10)\text{the}\text{value}\text{of}S=-14\\ \text{and}\text{at}(6,0),\text{we}\text{get}\\ S=6\times 6-2\times 0\\ S=36-0\\ S=36\\ \therefore \text{at}(6,0)\text{the}\text{value}\text{of}S\text{is}36.\end{array}$

Hence, the maximum value is 36 attain at (6,0) and minimum value is $-38$ attain at $(-3,10).$