Question

A company produces mixers and food processors. Profit on selling one mixer and one food processor is Rs 2,000 and Rs 3,000 respectively. Both the products are processed through three machines A, B, C. The time required in hours for each product and total time available in hours per week on each machine arc as follows:

Machine	Mixer	Food Processor	Available time
A	3	3	36
B	5	2	50
C	2	6	60

How many mixers and food processors should be produced in order to maximize the profit?

Answer 1

Let x = number of mixers are sold
      y = number of food processors are sold

Profit function z = 2000x + 3000y

This is the objective function which is to be maximized. From the given table in the problem, the constraints are

3x + 3y ≤ 36        (above machine A)
5x 2y ≤ 50           ( about machine B)
2x + 6y ≤ 60        ( about machine C)

As the number of mixers and food processors are non-negative.

x ≥ 0, y ≥ 0
Mathematical model of L.P.P. is
Maximize Z = 2000x + 3000y  Subject to 
3x + 3y ≤ 36, 5x 2y ≤ 50, 2x + 6y ≤ 60
and x ≥ 0, y ≥ 0

First we draw the lines AB, CD and EF whose equations are 3x + 3y = 36, 5x + 2y = 50 and 2x + 6y = 60 respectively.

Line	Equation	Points on the X-axis	Points on the Y-axis	Sign	Region
AB	3x + 3y = 36	A(12,0)	B(0,12)	≤	origin side of line AB
CD	5x + 2y = 50	C(10,0)	D(0,25)	≤	origin side of line CD
EF	2x + 6y = 60	E(30,0)	F(0,10)	≤	origin side of line EF

The feasible region is OCPQFO which is shaded in the graph.

The vertices of the feasible region are O (0, 0), C (10, 0), P, Q and F (0, 10).

P is the point of intersection of the lines

3x + 3y = 36      ....(1)

and 5x + 2y = 50     ....(2)

Multiplying equation (1) by 2 and equation (2) by 3, we get,

6x + 6y = 72

15x + 6y = 150

On subtracting, we get

9x = 78

∴ x = `26/3`

∴ from (1), `3(26/3) + 3"y" = 36`

∴ 3y = 10

∴ y = `10/3`

∴ P = `(26/3,10/3)`

Q is the point of intersection of the lines

3x + 3y = 36     . ...(1)

and 2x + 6y = 60     ...(2)

Multiplying equation (1) by 2, we get

6x + 6y = 72

Subtracting equation (3), from this equation, we get

4x = 12

∴ x = 3

∴ from (1), 3(3) + 3y = 36

∴ 3y = 27

∴ y = 9

∴ Q = (3, 9)

The values of the objective function z = 2000x  + 3000y at these vertices are

z(O) = 2000(0) + 3000(0) = 0 + 0 = 0

z(C) = 2000(10) + 3000(0) = 20000 + 0 = 20000

z(P) = `2000(26/3) + 3000(10/3) = 52000/3 + 30000/3 = 82000/3`

z(Q) = 2000(3) + 3000(9) = 6000 + 27000 = 33000

z(F) = 2000(0) + 3000(10) = 30000 + 0 = 30000

∴ the maximum value of z is 33000 at the point (3, 9).

Hence, 3 mixers and 9 food processors should be produced in order to get the maximum profit of ₹ 33,000.