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Solve the following problem : A Company produces mixers and processors Profit on selling one mixer and one food processor is ₹ 2000 and ₹ 3000 respectively. Both the products are processed through

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Question

Solve the following problem :

A Company produces mixers and processors Profit on selling one mixer and one food processor is ₹ 2000 and ₹ 3000 respectively. Both the products are processed through three machines A, B, C The time required in hours by each product and total time available in hours per week on each machine are as follows:

Machine/Product Mixer per unit Food processor per unit Available time
A 3 3 36
B 5 2 50
C 2 6 60

How many mixers and food processors should be produced to maximize the profit?

Graph
Sum
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Solution

Let x mixers and y food processors be produced by the company.
∴ Total profit Z = 2000x + 3000y
This is the objective function to be maximized.
From the given information, the constraints are
3x + 3y ≤ 36, 5x + 2y ≤ 50, 2x + 6y ≤ 60, x ≥ 0, y ≥ 0
∴ Given problem can be formulated as
Maximize Z = 2000x + 3000y
Subject to, 3x + 3y ≤ 36, 5x + 2y ≤ 50, 2x + 6y ≤ 60, x ≥ 0, y ≥ 0
To draw the feasible region, construct table as follows:

Inequality 3x + 3y ≤ 36 5x + 2y ≤50 2x + 6y ≤ 60 
Corresponding equation (of line) 3x + 3y = 36 5x + 2y = 50 2x + 6y = 60
Intersection of line with X-axis (12, 0) (10, 0) (30, 0)
Intersection of line with Y-axis (0, 12) (0, 25) (0, 10)
Region Origin side Origin side Origin side

Shaded portion OABCD is the feasible region,
whose vertices are O ≡ (0, 0), A ≡ (10, 0), B, C and D ≡ (0, 10)
B is the point of intersection of the lines
3x + 3y = 36 i.e. x + y = 12 and 5x + 2y = 50
Solving the above equations, we get
B ≡ `(26/3, 10/3)`
C is the point of intersection of the lines 3x + 3y = 36
i.e. x + y = 12 and 2x + 6y = 60
i.e. x + 3y = 30
Solving the above equations, we get
C ≡ (3, 9)
Here the objective function is
Z = 2000x + 3000y
∴ Z at O(0, 0) = 2000(0) + 3000(0) = 0
Z at A(10, 0) = 2000(10) + 3000(0) = 20000
Z at B`(26/3, 10/3) = 2000(26/3) + 3000(10/3) = (82000)/(3)` = 27333.33
Z at C(3, 9) = 2000(3) + 3000(9) = 33000
Z at D(0, 10) = 2000(0) + 3000(10) = 30000
∴ Z has maximum value 33000 at C(3, 9).
∴ Z is maximum when x = 3, y = 9

Thus, the company should produce 3 mixers and 9 food processors to gain maximum profit of ₹ 33000.

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Chapter 6: Linear Programming - Miscellaneous Exercise 6 [Page 105]

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Balbharati Mathematics and Statistics 2 (Commerce) [English] Standard 12 Maharashtra State Board
Chapter 6 Linear Programming
Miscellaneous Exercise 6 | Q 4.11 | Page 105
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