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प्रश्न
Minimize Z = 2x + 3y subject to constraints
x + y ≥ 6, 2x + y ≥ 7, x + 4y ≥ 8, x ≥ 0, y ≥ 0
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उत्तर
To draw the feasible region, construct table as follows:
| Inequality | x + y ≥ 6 | 2x + y ≥ 7 | x + 4y ≥ 8 |
| Corresponding equation (of line) | x + y = 6 | 2x + y = 7 | x + 4y = 8 |
| Intersection of line with X-axis | (6, 0) | `(7/2, 0)` | (8, 0) |
| Intersection of line with Y-axis | (0, 6) | (0, 7) | (0, 2) |
| Region | Non-origin side | Non-origin side | Non-origin side |
Shaded portion XABCDY is the feasible region, whose vertices are A(8, 0), B, C and D(0, 7).
B is the point of intersection of x + 4y = 8 and x + y = 6.
Solving the above equations, we get
x = `16/3`, y = `2/3`
∴ B ≡ `(16/3, 2/3)`
C is the point of intersection of 2x + y = 7 and x + y = 6.
Solving the above equations, we get
x = 1, y = 5
∴ C ≡ (1, 5)
Here, the objective function is Z = 2x + 3y
∴ Z at A(8, 0) = 2(8) + 3(0) = 16
Z at B`(16/3, 2/3)`
= `2(16/3) + 3(2/3)`
= `32/3 + 6/3`
= `38/3`
Z at C(1, 5) = 2(1) + 3(5) = 17
Z at D(0, 7) = 2(0) + 3(7) = 21
∴ Z has minimum value `38/3` at x = `16/3` and y = `2/3`.
संबंधित प्रश्न
A firm manufactures two products, each of which must be processed through two departments, 1 and 2. The hourly requirements per unit for each product in each department, the weekly capacities in each department, selling price per unit, labour cost per unit, and raw material cost per unit are summarized as follows:
| Product A | Product B | Weekly capacity | |
| Department 1 | 3 | 2 | 130 |
| Department 2 | 4 | 6 | 260 |
| Selling price per unit | ₹ 25 | ₹ 30 | |
| Labour cost per unit | ₹ 16 | ₹ 20 | |
| Raw material cost per unit | ₹ 4 | ₹ 4 |
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Minimize Z = 4x + 5y
Subject to the constraints 5x + y ≥ 10, x + y ≥ 6, x + 4y ≥ 12, x, y ≥ 0
Solution: Convert the constraints into equations and find the intercept made by each one of it.
| 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 |
∵ Origin has not satisfied the inequations.
∴ Solution of the inequations is away from origin.
The feasible region is unbounded area which is satisfied by all constraints.
In the figure, ABCD represents
The set of the feasible solution where
A(12, 0), B( ___, ___ ), C ( ___, ___ ) and D(0, 10).
The coordinates of B are obtained by solving equations
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 ___
