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A Firm Manufacturing Two Types of Electric Items, a and B, Can Make a Profit of Rs 20 per Unit of a and Rs 30 per Unit of B. - Mathematics

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Sum

A firm manufacturing two types of electric items, A and B, can make a profit of Rs 20 per unit of A and Rs 30 per unit of B. Each unit of A requires 3 motors and 4 transformers and each unit of B requires 2 motors and 4 transformers. The total supply of these per month is restricted to 210 motors and 300 transformers. Type B is an export model requiring a voltage stabilizer which has a supply restricted to 65 units per month. Formulate the linear programing problem for maximum profit and solve it graphically.

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Solution

Let units of item A and y units of item B were manufactured.
Number of items cannot be negative.
Therefore, \[x, y \geq 0\]

The given information can be tabulated as follows:

Product        Motors Transformers
A(x) 3 4
B(y) 2 4
Availability 210 300


Further, it is given that type B is an export model, whose supply is restricted to 65 units per month.
Therefore, the constraints are

\[3x + 2y \leq 210\]
\[4x + 4y \leq 300\]
\[y \leq 65\]

A and B can make a profit of Rs 20 per unit of A and Rs 30 per unit of B.Therefore, profit gained from x units of item A and y units of item B is Rs 20x and Rs 30y respectively.
Total profit = Z = \[20x + 30y\] which is to be maximised.
Thus, the mathematical formulat‚Äčion of the given linear programmimg problem is 
Max Z =  \[20x + 30y\]

subject to

\[3x + 2y \leq 210\]
\[4x + 4y \leq 300\]
\[y \leq 65\]

\[x, y \geq 0\]

First we will convert inequations into equations as follows:
3x + 2y = 210, 4x + 4y = 300, y = 65, x = 0 and y = 0

Region represented by 3x + 2y ≤ 210:
The line 3x + 2y = 210 meets the coordinate axes at A1(70, 0) and B1(0, 105) respectively. By joining these points we obtain the line 3x + 2y = 210.  Clearly (0,0) satisfies the 3x + 2y = 210. So,the region which contains the origin represents the solution set of the inequation 3x + 2y ≤ 210.

Region represented by 4x + 4y ≤ 300:
The line 4x + 4y = 300 meets the coordinate axes at C1(75, 0) and D1(0, 75) respectively. By joining these points we obtain the line
4x + 4y = 300.Clearly (0,0) satisfies the inequation 4x + 4y ≤ 300. So,the region which contains the origin represents the solution set of the inequation
4x + 4y ≤ 300.

y = 65 is the line passing through the point E1(0, 65) and is parallel to X axis.

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 ≥ 0.
The feasible region determined by the system of constraints 3x + 2y ≤ 210, 4x + 4y ≤ 300, y ≤ 65, x ≥ 0, and y ≥ 0 are as follows

The corner points are O(0, 0),  E1(0, 65), G1

\[\left( 10, 65 \right)\], F1(60, 15) and A1(70, 0).


The values of Z at these corner points are as follows
 

Corner point Z= 20x + 30y
O 0
E1 1950
G1 2150
F1 1650
A1 1400

The maximum value of Z is 2150 which is attained at G1
\[\left( 10, 65 \right)\] Thus, the maximum profit is Rs 2150 obtained when 10 units of item A and 65 units of item B were manufactured.
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 13 | Page 52

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