A merchant plans to sell two types of personal computers — a desktop model and a portable model that will cost 25,000 and 40,000 respectively. He estimates that the total monthly demand for computers will not exceed 250 units. Determine the number of units of each type of computers which the merchant should stock to get maximum profit if he does not want to invest more than ` 70 lakhs and his profit on the desktop model is ` 4,500 and on the portable model is 5,000. Make an L.P.P. and solve it graphically.

Let x be the units be desktop demanded, and y be the units of

Portable model demanded.

According to a question we have the following objective function

and constraints.

Here the objective is to maximise profit.

Objective function: z = 4500x + 2500y

Constraint:

x + y ≤ 250

25000x + 40000y ≤ 70,00,000

⇒ 5x + 8y ≤ 1400

x ≥ 0, y ≥ 0

The maximum value of z can only be obtained at the corner points of the feasible region. So we need to check the value of z at all corner points of the feasible region.

So, first, we will be finding out the feasible region by drawing the regions defined by constraints.

For plotting feasible region, we will be using the fundamentals of a straight line to get the feasible region as shown in the figure.

Clearly ABCD represents the feasible region and corner points are determined by solving:

x + y = 250 and 5x + 8y = 1400

x = 0 and 5x+8y = 1400

y = 0 and x + y = 250

& x = 0 and y = 0

To solve -

∴Value of objective function z

at point A = 4500(250) + 2500(0) = 1125000

Value of Z at point B = 4500×(200) + 2500(50) = 1150000

Value of Z at point C = 4500(0) + 2500(175) = 875000

Value of Z at point D = 4500×0 + 2500× 0 = 0

Z is maximum at point C (200,50)

∴ Profit will be maximised for 200 units of x and 50 units of y and the maximum value of Z = maximum profit = Rs. 1150000