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Solve the following problem : A person makes two types of gift items A and B requiring the services of a cutter and a finisher. Gift item A requires 4 hours of cutter's time and 2 hours of finisher' - Mathematics and Statistics

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Question

Solve the following problem :

A person makes two types of gift items A and B requiring the services of a cutter and a finisher. Gift item A requires 4 hours of cutter's time and 2 hours of finisher's time. B requires 2 hours of cutters time, 4 hours of finishers time. The cutter and finisher have 208 hours and 152 hours available times respectively every month. The profit of one gift item of type A is ₹ 75 and on gift item B is ₹ 125. Assuming that the person can sell all the items produced, determine how many gift items of each type should be make every month to obtain the best returns?

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Solution

Let x gift items of type A and y gift items of type B be produced by the person.
∴ Total profit Z = 75x + 125y
This is the objective function to be maximized.
The given information can be tabulated as shown below:

  Type A (x) Type B (y) Total time available (in hours)
Cutter 4 2 208
Finisher 2 4 152

∴ The constraints are 4x + 2y ≤ 208, 2x + 4y ≤ 152, x ≥ 0, y ≥ 0
∴ Given problem can be formulated as
Maximize Z = 75x + 125y
Subject to, 4x + 2y ≤ 208, 2x + 4y ≤ 152, x ≥ 0, y ≥ 0
To draw feasible region, construct table as follows:

Inequality 4x + 2y ≤ 208 2x + 4y ≤ 152
Corresponding equation (of line) 4x + 2y = 208 2x + 4y = 152
Intersection of line with X-axis (52, 0) (76, 0)
Intersection of line with Y-axis (0, 104) (0, 38)
Region Origin side Origin side

Shaded portion OABC is the feasible region,
whose vertices are O ≡ (0, 0),
A ≡ (52, 0), B and C ≡ (0, 38).
B is the point of intersection of the lines 4x + 2y = 208 i.e. 2x + y = 104 and 2x + 4y = 152

Solving the above equations, we get B ≡ (44, 16)
Here, the objective function is Z = 75x + 125y
∴ Z at O(0, 0) = 75(0) + 125(0) = 0
Z at A(52, 0) = 75(52) + 125(0) = 3900
Z at B(44, 16) = 75(44) + 125(16) = 5300
Z at C(0, 38) = 75(0) + 125(38) = 4750
∴ Z has maximum value 5300 at B(44, 16)
∴ Z is maximum, when x = 44, y = 16
Thus, a person should make 44 gift items of type A and 16 gift items of type B every month to obtain the best returns of ₹ 5300.

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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 4.13 | Page 105

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