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Question

The minimum value of 3x + 5y such that:

3x+5y15

4x+9y8

13x+2y2

x0, y0

is ........................

  1. 0

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Solution

The correct option is A 0


Z = 3x + 5y

Let us consider 3x+5y15 ...(i)

If 3x + 5y = 15

x = 0, y = 3

x = 5, y = 0

Let us consider 4x+9y8 ...(ii)

x=0, y=89

x = 2, y = 0

Let us consider 13x+2y2 ...(iii)

If 13x + 2y = 2

x = 0, y = 1

x=213, y=0

Comparing (ii) and (iii)

4x + 9y = 8 ...(ii)

13x + 2y = 2 ... (iii)

From (ii) x=89y4

Putting in (iii)

13(89y)4+2y=2

104 - 117y + 8y = 8

109y = 96

y = 0.88

Hence, x = 0.018

Checking at corner points:

Z(0,89)=3(0)+5(89)=4.44

Z(0.018, 0.88) = 3(0.18) + 5(0.88) = 4.454

Z(213,0)=3(213)+5(0)=0.46

At (0, 0) Z = 0, so minimum value will be 0.

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