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Question
Solve the following problem :
Maximize Z = 5x1 + 6x2 Subject to 2x1 + 3x2 ≤ 18, 2x1 + x2 ≤ 12, x ≥ 0, x2 ≥ 0
Solution
To find the graphical solution, construct the table as follows:
| Inequation | equation | Double intercept form | Points (x1, x2) | Region |
| 2x1 + 3x2 ≤ 18 | 2x1 + 3x2 = 18 | `x_1/(9) + x_2/(6)` = 1 | A (9, 0) B (0, 6) |
2(0) + 3(0) ≤ 18 ∴ 0 ≤ 18 ∴ Origin-side |
| 2x1 + x2 ≤ 12 | 2x1 + x2 = 12 | `x_1/(6) + x_2/(12)` = 1 | C (6, 0) D (0, 12) |
2(0) + 1(0) ≤ 12 ∴ 0 ≤ 12 ∴ Origin-side |
| x1 ≥ 0 | x1 = 0 | – | – | R.H.S. of Y-axis |
| x2 ≥ 0 | x2 ≥ 0 | – | – | above X-axis |
The shaded portion OBEC is the feasible region.
Whose vertices are O (0, 0), B (0, 6), E, C (6, 0)
E is the point of intersection of the lines
2x1 + x2 = 12 ...(i)
and 2x1 + 3x2 = 18 ...(ii)
∴ By (i) – (ii), we get
2x1 + x2 = 12
2x1 + 3x2 = 18
– – –
–2x2 = – 6
∴ x2 = `(-6)/(-2)` = 3
Substituting x2 = 3 in (i), we get
2x1 + 3 = 12
∴ 2x1 = 12 – 3
∴ 2x1 = 9
∴ x1 = `(9)/(2)` = 4.5
∴ E (4.5,3)
Here, the objective function is Z = 5x1 + 6x2
Now, we will find maximum value of Z as follows:
| Feasible points | The value of Z = 5x1 + 6x2 |
| O (0, 0) | Z = 5(0) + 6(0) = 0 |
| B (0, 6) | Z = 5(0) + 6(6) = 36 |
| E (4.5, 3) | Z = 5(4.5) + 6(3) = 22.5 + 18 = 40.5 |
| E (4.5, 3) | Z = 5(6) + 6(0) = 30 |
∴ Z has maximum value 40.5 at E(4.5, 3)
∴ Z is maximum, when x1 = 4.5, x2 = 3.
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| C | ( ___, ___ ) | 4( ___) + 5(___ ) | ______ | |
| D | (0, 10) | 4(0) + 5(10) | 50 |
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