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Solve the following LPP: Maximize z =60x + 50y subject to x + 2y ≤ 40, 3x + 2y ≤ 60, x ≥ 0, y ≥ 0. - Mathematics and Statistics

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Question

Solve the following LPP:

Maximize z =60x + 50y  subject to

x + 2y ≤ 40, 3x + 2y ≤ 60, x ≥ 0, y ≥ 0.

Graph
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Solution

We first draw the lines AB and CD whose equations are x + 2y = 40 and 3x + 2y = 60 respectively.

Line Equation Points on the X-axis Points on the Y-axis Sign Region
AB x + 2y = 40 A(40,0) B(0,20) origin side of line AB
CD 3x + 2y = 60 C(20,0) D(0,30) 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 (20, 0), P and B (0, 20).
P is the point of intersection of the lines.

3x + 2y = 60       ....(1)

and x + 2y = 40     .....(2)

On subtracting, we get

2x = 20      ∴ x = 10

Substituting x = 10 in (2), we get

10 + 2y = 40

∴ 2y = 30

∴ y = 15

∴ P is (10, 15)

The values of the objective function z = 60x + 50y at these vertices are

z(O) = 60(0) + 50(0) = 0 + 0 = 0

z(C) = 60(20) + 50(0) = 1200 + 0 = 1200

z(P) = 60(10) + 50(15) = 600 + 750 = 1350

z(B) = 60(0) + 50(20) = 0 + 1000 = 1000

∴ z has maximum value 1350 at x = 10, y = 15. 

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Linear Programming Problem (L.P.P.)
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Chapter 7: Linear Programming - Miscellaneous exercise 7 [Page 244]

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Balbharati Mathematics and Statistics 1 (Arts and Science) [English] Standard 12 Maharashtra State Board
Chapter 7 Linear Programming
Miscellaneous exercise 7 | Q II) 6) ii) | Page 244

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