Question

A merchant plans to sell two types of personal computers a desktop model and a portable model that will cost Rs 25000 and Rs 40000 respectively. He estimates that the total monthly demand of 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 Rs  70 lakhs and if his profit per unit on the desktop model is Rs  4500 and on protable model is Rs 5000.

Answer 1

Let the number of desktop model = x and the number of portable model = y.

∴ Objective function to earn maximum profit

Z = 4500x + 5000y

Constraints for the total number of computers is
x + y ≤ 250

∵ monthly demand of computers is not more than 250.

Total cost of computers is

25000x + 40000y ? 70,000,00

∴ x and y are number of computers.

∴ x ≥ 0 and y ≥ 0

Mathematically formulation of Linear Programming

Problem is as following :

Maximum Z = 4500x + 5000y

Subject to the constraints

x + y ≤ 250

25000x + 40,000y = 70,000,00

x ≥ 0 and y ≥ 0

Converting the given in equations, into the equations

x + y = 250 …..(1)

25000x + 40000y = 7000000

⇒ 25x + 40y = 7000 …..(2)

Region represented by x + y ≤ 250 : 

The line x + y = 250 meets the coordinate axis at the points A(250, 0) and B(0, 250).
x + y = 250

x	250	0
y	0	250

A(250, 0); B(0, 250)

Join point A to B to obtain the line. 

Clearly (0,0) satisfies the in equation 0 + 0 = 0 ≤ 250. 

So the region containing origin represents the solution set of the in equation.

Region represented by 25x + 40y ≤ 7000 : 

The line 25x + 40y = 7000 meets the coordinate axis on the points C(280, 0) and D(0, 175).

x	280	0
y	0	175

C(280, 0);D(0, 175)

Join point C to D to obtain the line. 

Clearly (0,0) satisfies the given in equation 25(0) + 40(0) = 0 ≤ 7000. 

So the region containing the origin represents the solution set of the in equation.

Region represented by x ≥ 0, y ≥ 0 : 

Since every point in the first quadrant satisfies these inequations. 

So the first quadrant is the region represented by the inequations x ≥ 0 and y ≥ 0.

The coordinates of the point of intersection of lines x + y = 250 and 25x + 40y = 7000 are x = 200 and y = 50.
The shaded region OAED is the common region of the above in equations. 

This region is the feasible solution of the Linear Programming Problem.

The coordinates of the comer points of this region are O(0, 0), A(250, 0), E(200, 50) and D(0, 175).

The value of objective function on these points is given in the following table :

Point	x-coordinate	y-coordinate	Objective function Z = 4500x + 5000y
O	0	0	ZG = 4500 x 0 + 5000 x 0 = 0
A	250	0	ZA = 4500 x 250 + 5000 x 0 = 11,25,000
E	200	50	ZE = 4500 x 200 + 5000 x 50 = 900000 + 250000 = 1150000
D	0	175	ZD = 4500 x 0 + 5000 x 175 = 8,75000

It is clear from the table that the objective function value is maximum at point E(200, 50)
and z = 11,50,000
So the merchant should purchase 200 desktop computer and 50 portable computer to earn maximum profit of Rs 11,50,000.