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प्रश्न
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?
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उत्तर
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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| Inequations | Equations | X intercept | Y intercept | Region |
| 5x + y ≥ 10 | 5x + y = 10 | ( ___, 0) | (0, 10) | Away from origin |
| x + y ≥ 6 | x + y = 6 | (6, 0) | (0, ___ ) | Away from origin |
| x + 4y ≥ 12 | x + 4y = 12 | (12, 0) | (0, 3) | Away from origin |
| x, y ≥ 0 | x = 0, y = 0 | x = 0 | y = 0 | 1st quadrant |
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In the figure, ABCD represents
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A(12, 0), B( ___, ___ ), C ( ___, ___ ) and D(0, 10).
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x + 4y = 12 and x + y = 6
The coordinates of C are obtained by solving equations
5x + y = 10 and x + y = 6
Hence the optimum solution lies at the extreme points.
The optimal solution is in the following table:
| Point | Coordinates | Z = 4x + 5y | Values | Remark |
| A | (12, 0) | 4(12) + 5(0) | 48 | |
| B | ( ___, ___ ) | 4( ___) + 5(___ ) | ______ | ______ |
| C | ( ___, ___ ) | 4( ___) + 5(___ ) | ______ | |
| D | (0, 10) | 4(0) + 5(10) | 50 |
∴ Z is minimum at ___ ( ___, ___ ) with the value ___
If z = 200x + 500y .....(i)
Subject to the constraints:
x + 2y ≥ 10 .......(ii)
3x + 4y ≤ 24 ......(iii)
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