# 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. - Mathematics

Sum

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 be the number of items of 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.

∴ xy ≤ 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 items of A and 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 + = 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 $\frac{1}{2}$ y ​ ≤ 16:
The line ​x +  $\frac{1}{2}$ = 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$\frac{1}{2}$ = 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 ≥ 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 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 = 8 and = 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.

Concept: Graphical Method of Solving Linear Programming Problems
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#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 30 Linear programming
Exercise 30.4 | Q 32 | Page 54