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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.
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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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