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
∴ 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
∴ 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.