A small firm manufacturers items A and B. The total number of items A and B that it can manufacture in a day is at the most 24. Item A takes one hour to make while item B takes only half an hour. The maximum time available per day is 16 hours. If the profit on one unit of item A be Rs 300 and one unit of item B be Rs 160, how many of each type of item be produced to maximize the profit? Solve the problem graphically.
Solution
Let the firm manufacture x items of A and y items of B per day.
Number of items cannot be negative.
Therefore, x ≥0 and y ≥0.
It is given that the total number of items manufactured per day is at most 24.
∴ x + y ≤ 24
Item A takes 1 hour to make and item B takes 0.5 hour to make.
The maximum number of hours available per day is 16 hours.
∴ x + 0.5y ≤ 16
If the profit on one unit of item A be Rs 300 and one unit of item B be Rs 160.Therefore, profit gained on x items of A and y items of B is
Rs 300x and Rs 160y respectively.
∴ Total profit, Z = 300x + 160y
Thus, the mathematical formulation of the given problem is:
Maximise Z = 300x + 160y
subject to the constraints
x+ y ≤ 24
x + 0.5y ≤ 16
x, y ≥0
First we will convert inequations into equations as follows:
x + y = 24, x + 0.5y = 16, x = 0 and y = 0
Region represented by x + y ≤ 24:
The line x + y = 24 meets the coordinate axes at A1(24, 0) and B1(0, 24) respectively. By joining these points we obtain the line x + y = 24. Clearly (0,0) satisfies the x + y = 24. So, the region which contains the origin represents the solution set of the inequation
x + y ≤ 24.
Region represented by x + 0.5y ≤ 16:
The line x + 0.5y = 16 meets the coordinate axes at C1(16, 0) and D1(0, 32) respectively. By joining these points we obtain the line x + 0.5y = 16. Clearly (0,0) satisfies the inequation x + 0.5y ≤ 16. So,the region which contains the origin represents the solution set of the inequation x + 0.5y ≤ 16.
Region represented by x ≥ 0 and 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 feasible region determined by the system of constraints x + y ≤ 24, x + 0.5y ≤ 16, x ≥ 0and y ≥ 0 are as follows.
The corner points are O(0, 0) ,C1(16, 0), E1(8, 16) and B1(0, 24).
The value of Z at these corner points are as follows:
| Corner point | Z = 300x + 160y |
| O(0, 0) | 0 |
| C1(16, 0) | 4800 |
| E1(8, 16) | 4960 |
| B1(0, 24) | 3840 |
Thus, the maximum value of Z is 4960 at E1(8, 16).
Thus, 8 units of item A and 16 units of item B should be manufactured per day to maximise the profits.
