Question

LP 3. A manufacturer wants to maximize the profit of two products. Product l yields aprcfit of $ 1.50 per unit, and product Il yields a profit of $2.00 per unit. Market tests and available resources have indicated the following constraints The combined production level should not exceed 1200 units per month. . The demand for product Il is no more than half the demand for product . The production level of product l is less than or equal to 600 units plus three times the production level of product ll.

Answer 1

Let X and Y be the quantity of Product I and Product II respectively.

The objective is to maximize the profit. Hence the objective function Z, is given as below

Z = 1.5X+2Y

As total production cannot exceed 1200, we have the below constraint

X+Y1200

As demand for product II cannot be more than half of demand for product I, we have the below constraint

Y0.5X

-0.5X+Y0

As production level of product I is less than or equal to three time that of product II plus 600 we have the below constraint

X 3Y+600

X-3Y 600

For non negativity, we also have the constraint that X,Y 0

To find the solution, we can find plot the above constraint on the graph and the find the feasible solution area. We then find the value of the objective function at the corner points of the feasible solution. The point where the objective function has the maximum value is the solution.

The graph of the equation and the feasible area is shown in the below image

We can see that there are 4 corner points of the feasible solution.

A (0,0), B(800,400) C(1050,150) D(600,0)

We find the value of objective function 1.5X+2Y at each of the corner points

At A, the value is 1.5*0+2*0 = 0

At B, the value is 1.5*800+2*400 = 2000

At C, the value is 1.5*1050+2*150 =1875

At A, the value is 1.5*600+2*0 = 900

We can see that the maximum value occurs are B(800,400). Hence 800 Product I and 400 Product II need to be produced to maximize the profit.

The maximum profit is the value of objective function which is $2000