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For Positive Integers X And Y; 3x + 8y = 68. Find The Maximum Value Of x^2*y^3.

Sujoy Das

07/02/2018 2 0

Question:

For positive integers x and y; 3x + 8y = 68. Find the maximum value of x^2*y^3.

Answer:

Method 1:

3x +8y = 68; x = (68-8y)/3

i.e for y = 1; x = 20

Hence y's values will increase in step of 3 and x's value will decrease in steps of 8.

So, possible positive solutions are:

x y

20 1

12 4

4 7

Max value at 12, 4 = 9216.

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