In: Operations Management
Aspen carpentry makes bookcases and desks. Each bookcase requires 5 hours of . woodworking and 3 hours of finishing. Each desk requires 10 hours of woodworking and 3 hours of finishing. Each month the shop has 600 hours of labor available f~r woodworking and 240 hours available for finishing,. The profit on each bookcase is $75 and on each desk is $140. How many of each product should be made each month In order to maximize profit? What is the max1mum profit?
We can solve this problem using graphical method of linear programming. It is given that Aspen carpentry makes bookcases and desks. The profit on each bookcase is $75 and on each desk is $140 & we have to find the maximum profit. Let x denote the bookcases & y represents desks. We are given these constraints: Each bookcase requires 5 hours of woodworking and 3 hours of finishing & each desk requires 10 hours of woodworking and 3 hours of finishing.
Also, it is given that each month the shop has 600 hours of labor available for woodworking and 240 hours available for finishing.
So, we can write the problem as:
Subject to,
To draw the constraints we will change the inequality into equality,
5x+10y=600
x | 0 | 120 |
y | 60 | 0 |
3x+3y=240
x | 0 | 80 |
y | 80 | 0 |
With the help of the above mentioned graph we can see that the extreme point coordinates are 0(0,0), A(0,60), B(40,40) & C(80,0). Now, we will find the value of the objective function at each of these extreme points.
Extreme Point Coordinates (x,y) | Value of Objective function Z=75x+140y |
0(0,0) | 75(0)+140(0)=0 |
A(0,60) | 75(0)+140(60)=8400 |
B(40,40) | 75(40)+140(40)=8600 |
C(80,0) | 75(80)+140(0)=6000 |
In the above table, we can see that the value of the objective function is maximum at the extreme point B(40,40). So, the optimal value of x=40 & y=40 & the maximum profit is 8,600.
So, in order to maximize profit, Aspen carpentry should make 40 bookcases & 40 desks each month & the maximum profit that it will earn will be $8,600.