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प्रश्न
Solve the following problem:
Maximize Z = 4x1 + 3x2 Subject to 3x1 + x2 ≤ 15, 3x1 + 4x2 ≤ 24, x1 ≥ 0, x2 ≥ 0
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उत्तर
To find the graphical solution, construct the table as follows:
| Inequation | Equation | Double intercept form | Points (x1, x2) | Region |
| 3x1 + x2 ≤ 15 | 3x1 + x2 = 15 | `x_1/(5) + x_2/(15)` = 1 | A (5, 0) B (0, 15) |
3(0) + 0 ≤ 15 ∴ 0 ≤ 15 ∴ origin side |
| 3x1 + 4x2 ≤ 24 | 3x1 + 4x2 = 24 | `x_1/8 + x_2/(6)` = 1 | C (8, 0) D (0, 6) |
3(0 + 4(0) ≤ 24 ∴ 0 ≤ 24 ∴ origin side |
| x1 ≥ 0 | x1 = 0 | – | – | R.H.S. of Y-axis |
| x2 ≥ 0 | x2 = 0 | – | – | above X-axis |
Shaded portion ODEA is the feasible region.
Whose vertices are O (0, 0), D (0, 6), E, A (5, 0)
E is the point of intersection of the lines
3x1 + x2 = 15 …(i)
and 3x1 + 4x2 = 24 …(ii)
∴ By (i) – (ii), we get
3x1 + x2 = 15
3x1 + 4x2 = 24
– – –
–3x2 = – 9
∴ x2 = `(-9)/(-3)`
∴ x2 = 3
Substituting x2 = 3 in i, we get
3x1 + 3 = 15
∴ 3x1 = 15 – 3
∴ 3x1 = 12
∴ x1 = `(12)/(3)` = 4
∴ E (4, 3)
Here, the objective function is Z = 4x1 + 3x2
Now, we will find maximum value of Z as follows:
| Feasible points | The value of Z = 4x1 + 3x2 |
| O (0, 0) | Z = 4(0) + 3(0) = 0 |
| D (0, 6) | Z = 4(0) + 3(0) = 0 |
| E (4, 3) | Z = 4(4) + 3(3) = 16 + 9 = 25 |
| A (5, 0) | Z = 4(4) + 3(3) = 16 + 9 = 25 |
∴ Z has maximum value 25 at E (4, 3)
∴ Z is maximum, when x1 = 4, x2 = 3.
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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 ___
