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 craftman’s time.
(i) What number of rackets and bats must be made if the factory is to work at full capacity?
(ii) If the profit on a racket and on a bat is Rs 20 and Rs 10 respectively, find the maximum profit of the factory when it works at full capacity.
Table:
Mathematical formulation of the given problem is as follows:
Maximise Z = x + y, P = 20 x + 10 y
subject to constrains
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Now we draw the graph of x + 2y = 28
For a = 0, 2 y = 28 or y = 14
For y = 0, x = 28
∴ line meets OX in A (28, 0) and OY in L(0, 14)
Again we draw the graph of 3x + y = 24
For x = 0, y = 24
For y = 0, 3x = 24 or x = 8
∴ line meets OX in B(8, 0) and OY in M(0, 24).
Since feasible region satisfies all the constraints.
∴ OBCL is the feasible region.
The comer points are O(0, 0), B(8, 0), C(4, 12), L(0, 14)
(i) For Z
At O(0, 0), Z = 0 + 0 = 0
At B(8, 0), Z = 8 + 0 = 8
At C(4, 12), Z = 4 + 12 = 16
At L(0, 14), Z = 0 + 14 = 14
∴ maximum value of Z is 16 at (4, 12)
∴ 4 tennis rackets and 12 cricket bats are made by the factory to run at full capacity.
(ii) For P
At O(0, 0), P = 0 + 0 = 0
At B(8, 0), P = 160 + 0 = 160
At C(4, 12), P = 80 + 120 = 200
At L(0, 14), P = 0 + 140 = 140
∴ maximum value of P is 200 at (4, 12)
∴ maximum profit = Rs. 200.
