A firm makes items A and B and the total number of items it can make in a day is 24. It takes one hour to make an item of A and half an hour to make an item of B. The maximum time available per day is 16 hours. The profit on an item of A is Rs 300 and on one item of B is Rs 160. How many items of each type should be produced to maximize the profit? Solve the problem graphically.
Solution
Let x and y be the number of items of A and B that should be produced each day to maximize the profit.
Number of items cannot be negative.
Therefore, x ≥ 0 , y ≥ 0
It is also given that the firm can produce at most 24 items in a day.
∴ x+ y ≤ 24
Also, the time required to make an item of A is one hour and time required to make an item of B is half an hour.
Therefore, the time required to produce x items of A and y items of B is `x + 1/2 y` hours. However, the maximum time available in a day is 16 hours. `x + 1/2 y ≤ 16`
It is given that the profit on an item of A is Rs 300 and on one item of B is Rs 160. Therefore, the profit gained from x items of A and y items of B is Rs 300x and Rs 160y respectively.
Total profit Z = 300x + 160y
The mathematical form of the given LPP is:
Maximize Z = 300x + 160y
subject to constraints:
\[x + y \leq 24\]
\[x + \frac{1}{2}y \leq 16\]
\[x \geq 0, y \geq 0\]
First we will convert inequations into equations as follows:
x + y = 24, x + \[\frac{1}{2}\] y = 16, 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 + \[\frac{1}{2}\] y ≤ 16, x ≥ 0 and y ≥ 0 are as follows.
he feasible region is shown in the figure
GRAPH
In the above graph, the shaded region is the feasible region. The corner points are O(0, 0), C_{1}(16, 0), E_{1}(8, 16), and B_{1}(0, 24).
The values of the objective function Z at corner points of the feasible region are given in the following table:
Corner Points 
Z = 300x + 160y 

O(0, 0) 
0 

C_{1}(16, 0) 
4800 

E_{1}(8, 16) 
4960 
← Maximum 
B_{1}(0, 24) 
3840 
Clearly, Z is maximum at x = 8 and y = 16 and the maximum value of Z at this point is 4960.
Thus, 8 items of A and 16 items of B should be produced in order to maximize the profit and the maximum profit is Rs 4960.