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Linear Programming

Question
CBSEENMA12033555

A company manufactures two types of novelty souvenirs made of plywood. Souvenirs of type A require 5 minutes each for cutting and 10 minutes each for assembling. Souvenirs of type B require 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours 20 minutes available for cutting and 4 hours tor assembling. The profit is Rs 5 each for type A and Rs 6 each for type B souvenirs. How many souvenirs of each type should be the company manufacture in order to maximise the profit?

Solution

Let the company manufacture x Souvenirs of type A and y Souvenirs of type B.
Let P be the profit.
Table

Type

Number

Time of cutting (minutes)

Time of assembling (minutes)

Profit (Rs.)

A

x

5x

10x

5x

B

y

8y

8y

6y

Total

 

5x + 8y

10x + 8y

5x + 6

We are to maximise
P = 5x + 6y
subject to constraints
5x + 8y ≤ 200
10x + 8y ≤ 240 or 5x + 4y ≤ 120
x ≥ 0, y ≥ 0
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
First we draw the graph of 5x + 8y = 200
For x = 0, 8y = 200 or y = 25
For y = 0, 5 x = 200 or x = 40
∴  line meets OX in A(40, 0) and OY in L(0, 25)
Again we draw the graph of 5x + 4y = 120
For x = 0, 4y = 120 or y = 30
For y = 0, 5x = 120 or x = 24
∴ line meets OX in B(24, 0) and OY in M(0. 30).

Since feasible region satisfies all the constraints.
∴ OBCL is the feasible region.
The comer points are O(0, 0), B(24, 0), C(8, 20), L(0, 25)
At O(0, 0), P = 5 × 0 + 6 × 0 = 0 + 0 = 0
At B(24, 0), P = 5 × 24 + 6 × 0 = 120 + 0 = 120
At C(8, 20), P = 5 × 8 + 6 × 20 = 40 + 120 = 160
At L(0, 25) P = 5 × 0 + 6 × 25 = 0 + 150 = 150
∴ maximum value = 160 at (8, 20)
∴ 8 souvenirs of type A and 20 souvenirs of type B are manufactured for maximum profit of Rs. 160.