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Question
Solve the following problem :
Minimize Z = 2x + 3y Subject to x – y ≤ 1, x + y ≥ 3, x ≥ 0, y ≥ 0
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Solution
To find the graphical solution, construct the table as follows:
| Inequation | Equation | Double intercept form | Points (x1, x2) | Region |
| x – y ≤ 1 | x – y = 1 | `x/(1) + y/(-1)` = 1 | A (1, 0) B (0, –1) |
0 – 0 ≤ 1 |
| x + y ≥ 1 | x + y = 3 | `x/(3) + y/(3)` = 1 | C (3, 0) D (0, 3) |
0 + 0 ≥ 3 ∴ 0 ≥ 3 ∴ non-origin side |
| x ≥ 0 | x = 0 | – | R.H.S. of Y- axis | |
| y ≥ | y = 0 | above X-axis |
The shaded portion Y' DE is the feasible region.
Whose vertices are D(0, 3) and E
E is the point of intersection of lines
x – y = 1 …(i)
x + y = 3 …(ii)
∴ By (i) + (ii), we get
x – y = 1
x + y = 3
2x = 4
∴ x = `(4)/(2)` = 2
Substituting x = 2 in (i), we get
2 – y = 1
∴ y = 1
∴ E(2, 1)
Here, the objective function is Z = 2x + 3y
Now, we will find minimum value of Z as follows:
| Feasible points | The value of Z = 2x + 3y |
| D(0, 3) | Z = 2(0) + 3(3) = 9 |
| E(2, 1) | Z = 2(2) + 3(1) = 4 + 3 = 7 |
∴ Z has minimum value 7 at E(2, 1)
∴ Z is minimum, when x = 2, y = 1.
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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 |
∵ 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 ___
Maximised value of z in z = 3x + 4y, subject to constraints : x + y ≤ 4, x ≥ 0. y ≥ 0
