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महाराष्ट्र राज्य शिक्षण मंडळएचएससी वाणिज्य (इंग्रजी माध्यम) इयत्ता १२ वी

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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प्रश्न

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?

आलेख
बेरीज

उत्तर

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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पाठ 6: Linear Programming - Miscellaneous Exercise 6 [पृष्ठ १०५]

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बालभारती Mathematics and Statistics 2 (Commerce) [English] 12 Standard HSC Maharashtra State Board
पाठ 6 Linear Programming
Miscellaneous Exercise 6 | Q 4.13 | पृष्ठ १०५

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