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Solve the following problem : Maximize Z = 5x1 + 6x2 Subject to 2x1 + 3x2 ≤ 18, 2x1 + x2 ≤ 12, x ≥ 0, x2 ≥ 0 - Mathematics and Statistics

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प्रश्न

Solve the following problem :

Maximize Z = 5x1 + 6x2 Subject to 2x1 + 3x2 ≤ 18, 2x1 + x2 ≤ 12, x ≥ 0, x2 ≥ 0

आलेख

उत्तर

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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अध्याय 6: Linear Programming - Miscellaneous Exercise 6 [पृष्ठ १०४]

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बालभारती Mathematics and Statistics 2 (Commerce) [English] 12 Standard HSC Maharashtra State Board
अध्याय 6 Linear Programming
Miscellaneous Exercise 6 | Q 4.01 | पृष्ठ १०४

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