Linear Programming 1
A company manufactures two types of components, motors and hubs for washing machines.A motor requires 60 hours to complete, while it takes a 20 hours for a hub to be made.
There are 1440 hours available for manufacture.
The number of motors and hubs manufactured must not exceed 48.
The profit on a motor is R300 while on a hub it is half of this amount.
Let the number of motors manufactured be x and the number of hubs made be y.
- Write down all the constraints.
- State the objective function.
- Graph the constraint equations.
- Determine, using the graph, the number of each component that must be made for maximum profit.
SOLUTION
1. Write down all the constraints.Take the information (in English) and convert to Mathematics as follows:
2. State the objective function.
This requires you to work with the profit and involve x and y in your answer.
This is done as follows:
P = 300x + 150y
3. Graph the constraint equations.
Represent the straight lines on one system of axes.
4. Determine, using the graph, the number of each component that must be made for maximum profit.
To get the possible values for maximum profit, we must substitute the co-ordinates of the vertices:
A(0; 24); B(36; 12); C(48; 0) and D(0; 0).
Upon substituting, the maximum value of profit occurs at vertex B.
P = 300x + 150y
=300(36) + 150(12)
=10 800 + 1 800
=R12 600