Question

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.

Answer 1

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

Region represented by x + y ≤ 24:

The line x + y = 24 meets the coordinate axes at A₁(24, 0) and \[B_1 \left( 0, 24 \right)\] 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 \[\frac{1}{2}\] y ​ ≤ 16:

The line ​x +  \[\frac{1}{2}\] y = 16 meets the coordinate axes at C₁(16, 0) and  \[D_1 \left( 0, 32 \right)\] respectively. By joining these points we obtain the line x + m\[\frac{1}{2}\] y = 16. Clearly, (0,0) satisfies the inequation x +  \[\frac{1}{2}\] y ​ ≤ 16. So,the region which contains the origin represents the solution set of the inequation x + \[\frac{1}{2}\] y ​ ≤ 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 +  \[\frac{1}{2}\] 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), C₁(16, 0),  E₁(8, 16), and B₁(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
C1(16, 0)	4800
E1(8, 16)	4960	← Maximum
B1(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.