Question

A factory makes tennis rackets and cricket bats. A tennis racket takes 1.5 hours of machine time and 3 hours of craftman’s time in its making while a cricket bat takes 3 hours of machine time and 1 hour of craftman’s time. In a day, the factory has the availability of not more than 42 hours of machine time and 24 hours of craftsman’s time. If the profit on a racket and on a bat is Rs 20 and Rs10 respectively, find the number of tennis rackets and cricket bats that the factory must manufacture to earn the maximum profit. Make it as an LPP and solve graphically.

Answer 1

Let the number of tennis rackets and cricket bats manufactured by factory be x and y respectively.

Here, profit on x rackets and y bats is the objective function Z.

Z = 20x + 10y …...(i)

We have to maximise Z subject to the constraints:

1.5x + 3y ≤ 42 …(ii) [Constraint for machine hour]

3x + y ≤ 24 …..(iii) [Constraint for craft man’s hour]

x, y ≥ 0 ..…(iv) [Non-negative constraints]

Graph of x = 0 and y = 0 is the y-axis and x-axis respectively.

∴ Graph of x ≥, y ≥ 0 is the Ist quadrant.

Graph of 1.5x + 3y = 42

x	0	28
y	14	0

∴ Graph for 1.5x + 3y ≤ 42 is the part of Ist quadrant which contains the origin.

Graph for 3x + y ≤ 24

Graph of 3x + y = 24

x	0	8
y	24	0

∴ Graph of 3x + y ≤ 24 is the part of Ist quadrant in which origin lie

Hence, shaded area OACB is the feasible region.

For coordinate of C equation 1. 5x + 3y = 42 and 3x + y = 24 are solved as

5x + 3y = 42 …(v)

3x + y = 24 …(vi)

⇒ y = 12 ⇒ x = 4 (Substituting y = 12 in (iv))

Now, value of objective function Z at each corner of feasible region is

Therefore, maximum profit is Rs 200, when factory make 4 tennis rackets and 12 cricket bats.