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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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