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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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