मराठी
महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता १२ वी

A company manufactures bicycles and tricycles each of which must be processed through machines A and B. Machine A has maximum of 120 hours available and machine B has maximum of 180 hours available. - Mathematics and Statistics

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प्रश्न

A company manufactures bicycles and tricycles each of which must be processed through machines A and B. Machine A has maximum of 120 hours available and machine B has maximum of 180 hours available. Manufacturing a bicycle requires 6 hours on machine A and 3 hours on machine B. Manufacturing a tricycle requires 4 hours on machine A and 10 hours on machine B.
If profits are Rs. 180 for a bicycle and Rs. 220 for a tricycle, formulate and solve the L.P.P. to determine the number of bicycles and tricycles that should be manufactured in order to maximize the profit.

आलेख
बेरीज

उत्तर

Let x number of bicycles and y number of tricycles be manufactured by the company.
Total profit Z = 180x + 220y
This is the objective function to be maximized. 
The given information can be tabulated as shown below:

  Bicycles (x) Tricycles (y) Maximum availability of time (hrs)
Machine A 6 4 120
Machine B 3 10 180

The constraints are 6x + 4y ≤ 120, 3x + 10y ≤ 180, x ≥ 0, y ≥ 0

Given problem can be formulated as 

Maximize Z = 180x + 220y

Subject to, 6x + 4y ≤ 120, 3x + 10y ≤ 180 , x ≥ 0, y ≥ 0.

To draw the feasible region, construct the table as follows:

Inequality 6x + 4y ≤ 120 3x + 10y ≤ 180
Corresponding equation (of line) 6x + 4y = 120 3x + 10y = 180
Intersection of line with X-axis (20, 0) (60, 0)
Intersection of line with Y-axis (0, 30) (0, 18)
Region Origin side Origin side

 

Shaded portion OABC is the feasible region, whose vertices are O=(0, 0), A =(20, 0), B and C = (0, 18)

B is the point of intersection of the lines 3x + 10y = 180 and 6x + 4y = 120.

Solving the above equations, we get
B = (10, 15) Here the objective function is,

Z = 180x + 220y

Z at O(0, 0) = 180(0) + 220(0) = 0
Z at A(20, 0) = 180(20) + 220(0) = 3600
Z at B(10, 15) = 180(10) + 220(15) = 5100
Z at C(0, 18) = 180(0) + 220(18) = 3960

Z has maximum value 5100 at B(10, 15)
Z is maximum when x = 10, y = 15

Thus, the company should manufacture 10 bicycles and 15 tricycles to gain maximum profit of Rs.5100.

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पाठ 7: Linear Programming - Miscellaneous exercise 7 [पृष्ठ २४४]

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बालभारती Mathematics and Statistics 1 (Arts and Science) [English] 12 Standard HSC Maharashtra State Board
पाठ 7 Linear Programming
Miscellaneous exercise 7 | Q 8) | पृष्ठ २४४

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