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Question
Solve the following problem :
Minimize Z = 4x + 2y Subject to 3x + y ≥ 27, x + y ≥ 21, 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) | Points (x1, x2) |
| 3x + y ≥ 27 | 3x + y = 27 | `x/(9) + y/(27)` = 1 | A (9, 0) B (0, 27) |
3(0) + 0 ≥ 27 |
| x + y ≥ 21 | x + y = 21 | `x/(21) + y/(21)` = 1 | C (21, 0) D (0, 21) |
(0) + 0 ≥ 21 ∴ 0 ≥ 21 ∴ non-origin-side |
| x ≥ 0 | x = 0 | – | R.H.S. of Y-axis | |
| y ≥ 0 | y = 0 | above X-axis |
The shaded portion CHB is the feasible region.
Whose vertices are C(21, 0), H and B(0, 27)
H is the point of intersection of lines
3x + y = 27 …(i)
x + y = 21 …(ii)
∴ By (i) – (ii), we get
3 x + y = 27
x + y = 21
– – –
2x = 6
∴ x = `(6)/(2)` = 3
Substituting x = 3 in (ii), we get
3 + y = 21
∴ y = 18
∴ H (3, 18)
Here, the objective function is Z = 4x + 2y
Now, we will find minimum value of Z as follows:
| Feasible points | The value of Z = 4x + 2y |
| C (21, 0) | Z = 4(21) + 2(0) = 84 |
| H (3, 18) | Z = 4(3) + 2(18) = 12 + 36 = 48 |
| B (0, 27) | Z = 4(0) + 2(27) = 54 |
∴ Z has minimum value 48 at H (3, 18)
∴ Z is minimum, when x = 3, y = 18
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