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Maximize Z = 60x + 50y Subject to x + 2y ≤ 40, 3x + 2y ≤ 60, x ≥ 0, y ≥ 0 - Mathematics and Statistics

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प्रश्न

Maximize Z = 60x + 50y Subject to x + 2y ≤ 40, 3x + 2y ≤ 60, x ≥ 0, y ≥ 0

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उत्तर

To find the graphical solution, construct the table as follows:

Inequation Equation Double intercept form Points (x, y) Region
x + 2y ≤ 40 x + 2y = 40 `x/(40) + y/(20)` = 1 A (40, 0)
B (0, 20)
0 + 2(0) ≤ 40
∴ 0 ≤ 40
∴ origin-side
3x + 2y ≤ 60 3x + 2y = 60 `x/(20) + y/(30)` = 1 C (20, 0)
D (0, 30)
3(0) + 2(0) ≤ 60
∴ 0 ≤  60
∴ origin-side
x ≥ 0 x = 0 R.H.S. of Y-axis
y ≥ 0 y = 0 Above X-axis

 

Shaded portion OBEC is the feasible region

Whose vertices are O (0, 0), B(0, 20), E and C (20, 0) 

E is the point of intersection of lines

x + 2y = 40        …(i)

3x + 2y = 60      …(ii)

∴ By (i) – (ii), we get

x + 2y = 40
3x + 2y = 60
  –  –      –       
–2x  = –20

∴ x = `(-20)/(-2)`

∴ x = 10

Substituting x = 10 in (i), we get 

10 + 2y = 40

∴ 2y = 40 – 10

∴ 2y = 30

∴ y = `(30)/(2)` = 15

∴ E = (10, 15)

Here, the objective function is Z = 60x + 50y 

Now, we will find maximum value of Z as follows:

Feasible points The value of Z = 60x + 50y
O (0, 0) Z = 60(0) + 50(0) = 0
B (0, 20) Z = 60(0) + 50(20) = 1000
E (10, 15) Z = 60(10) + 50(15) = 600 + 750 = 1350
C (20, 0) Z = 60(20) + 50(0) = 1200

∴ Z has maximum value 1350 at E (10, 15)

∴ Z is maximum, when x = 10, y = 15.

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

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बालभारती Mathematics and Statistics 2 (Commerce) [English] Standard 12 Maharashtra State Board
अध्याय 6 Linear Programming
Miscellaneous Exercise 6 | Q 4.06 | पृष्ठ १०४

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