Answer to Question #302896 in Operations Research for fiker

Question

A bed mart company is in the business of manufacturing beds and

pillows. The company has 40 hours for assembly and 32 hours for

finishing work per day. Manufacturing of a bed requires 4 hours for

assembly and 2 hours for finishing. Similarly a pillow requires 2 hours

for assembly and 4 hours for finishing. Profitability analysis indicates

that every bed would contribute Birr 80, while a pillow contribution is

Birr 55 respectively. Find out the daily production of the company to

maximise the contribution (profit). Solve the problem by graphical

method.

Answer 1

Let x, y denote the number of units of bed and pillow to be produced to maximize the profit.

The data given is tabulated below:

$\begin{array}{|c|c|c|c|} \hline & \text{Assembly} & \text{Finish} & \text{Profit} \\ \hline \text{Bed} & 4 & 2 & 80\\ \text{Pillow} & 2 & 4 & 55\\ \hline \text{Availability (Time)}& 40 & 32 & \\ \hline \end{array}$

The objective function is $z = 80x + 55y$. The constraints are on the available hours given by,

$\begin{aligned} 4x+2y &\le 40\\ 2x+4y &\le 32\\ x,y & \ge 0 \end{aligned}$

The linear programming model is,

$\text{Maximise } z = 80x+55y \\ \text{subject to}\\ \begin{aligned} 4x+2y &\le40\\ 2x+4y &\le 32\\ x,y &\ge 0 \end{aligned}$

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

$\begin{aligned} 4x+2y &=40 \qquad(1)\\ 2x+4y &= 32 \qquad(2)\\ \end{aligned}$

Equation (1) is the line passing through $(0,20)~ \& ~(10,0)$.

Equation (2) is the line passing through $(0,8)~ \& ~(16,0)$.

The graph is given below.

The feasible region is bounded by the constraints. The extreme points are $O(0,0), A(10,0), B(8,4), C(0,8)$. The optimal value occurs at one of the extreme points.

The point B is the point of intersection of the two lines which is obtained by solving the equations (1) and (2).

The value of z at these points are,

$z(O) = z(0,0) = 0\\ z(A) = z(10,0) = 800\\ z(B) = z(8,4) = 860\\ z(C) = z(0,8) = 440$

Hence the maximum value occurs at B(8,4) and the maximum value is z = 860.