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Solve the following LPP: Maximize z = 5x1 + 6x2 subject to 2x1 + 3x2 ≤ 18, 2x1 + x2 ≤ 12, x1 ≥ 0, x2 ≥ 0. - Mathematics and Statistics

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प्रश्न

Solve the following LPP:

Maximize z = 5x1 + 6x2 subject to 2x1 + 3x2 ≤ 18, 2x1 + x2 ≤ 12, x1 ≥ 0, x2 ≥ 0.

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उत्तर

First we draw the lines AB and CD whose equations are 2x1 + 3x2 = 18 and 2x1 + x2 = 12 respectively.

Line Equation Points on the X-axis Points on the Y-axis Sign Region
AB 2x1 + 3x2 = 18 A(9, 0) B(0, 6) origin side of line AB
CD 2x1 + x2 = 12 C(6, 0) O(0, 12) origin side of line CD

The feasible region is OCPBO which is shaded in the graph. The vertices of the feasible region are O (0, 0), C (6, 0), P and B (0, 6).

P is the point of intersection of the lines

2x1 + 3x2 = 18        ....(1)

and 2x1 + x2 = 12      ....(2)

On subtracting, we get

2x2 = 6

∴ x2 = 3

Substituting x2 = 3 in (2), we get

2x1 + 3 = 12

∴ x1 = 9

∴ P is `(9/2, 3)`

The values of objective function z = 5x1 + 6x2 at these vertices are

z(O) = 5(0) + 6(0) = 0 + 0 = 0

z(C) = 5(6) + 6(0) = 30 + 0 = 30

z(P) = `5(9/2) + 6(3) = 45/2 + 18 = (45 + 36)/2 = 81/2 = 40.5`

z(B) = 5(0) + 6(3) = 0 + 18 = 18

Maximum value of z is 40.5 when x1 = `9/2` y1 = 3.

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Linear Programming Problem (L.P.P.)
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 7: Linear Programming - Miscellaneous exercise 7 [पृष्ठ २४४]

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बालभारती Mathematics and Statistics 1 (Arts and Science) [English] Standard 12 Maharashtra State Board
पाठ 7 Linear Programming
Miscellaneous exercise 7 | Q II) 5) i) | पृष्ठ २४४

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