Question

Maximise 1170x1 + 1110x2

Subject to: 9x1 + 5x2 ≥ 500

7x1 + 9x2 ≥ 300

5x1 + 3x2 ≤ 1500

7x1 + 9x2 ≤ 1900

2x1 + 4x2 ≤ 1000

x1, x2 ≥ 0

-Find graphically the feasible region and the optimal solution.

Answer 1

Given,

$\text{Maximise}~ 1170x_{1} + 1110x_{2}\\ \text{subject to:}\\ \begin{aligned} 9x_1 + 5x_2 &≥ 500\\ 7x_1 + 9x_2 &≥ 300\\ 5x_1 + 3x_2 &≤ 1500\\ 7x_1 + 9x_2 &≤ 1900\\ 2x_1 + 4x_2 &≤ 1000\\ x_1, x_2 &≥ 0 \end{aligned}$

To solve it graphically, we consider the constraints as equations and draw straight lines.

$\begin{aligned} 9x_1 + 5x_2 &= 500~~\qquad(1)\\ 7x_1 + 9x_2 &= 300~~\qquad(2)\\ 5x_1 + 3x_2 &= 1500\qquad(3)\\ 7x_1 + 9x_2 &= 1900\qquad(4)\\ 2x_1 + 4x_2 &= 1000\qquad(5)\\ \end{aligned}$

The graph plotted is shown in the following figure.

The region of feasibility is bounded by the extreme points ABCD. The values of the objective function at the extreme points are given in the following table:

$\begin{array}{|c|c|} \hline \text{Extreme points}& \text{Value of ~} z = 1170x_1+1110x_2\\ \hline A(0,100) & 111000\\ &\\ B(500/9,0) & 65000\\ &\\ C(1900/7,0) & \dfrac{2223000}{7}\\ &\\ D(0,1900/9) & \dfrac{703000}{3}\\ &\\ \hline \end{array}$

Hence the maximum values occurs at C(1900/7,0) and the maximum value is $z = \dfrac{2223000}{7}$.