Numerical answer
If x and y are positive real numbers with x + y = 12, what is the maximum value of x²y?
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Answer: 256
Write 12 = x/2 + x/2 + y and apply AM ≥ GM to these three terms:
(x/2 + x/2 + y)/3 ≥ ∛(x²y/4).
So 4 ≥ ∛(x²y/4), giving x²y/4 ≤ 64 and x²y ≤ 256.
Equality holds when x/2 = y, i.e. x = 8 and y = 4. Then x²y = 64 × 4 = 256.
Tip: to maximise xᵐyⁿ with x + y fixed, split the total in the ratio m : n. Here 2 : 1 of 12 gives 8 and 4.
(x/2 + x/2 + y)/3 ≥ ∛(x²y/4).
So 4 ≥ ∛(x²y/4), giving x²y/4 ≤ 64 and x²y ≤ 256.
Equality holds when x/2 = y, i.e. x = 8 and y = 4. Then x²y = 64 × 4 = 256.
Tip: to maximise xᵐyⁿ with x + y fixed, split the total in the ratio m : n. Here 2 : 1 of 12 gives 8 and 4.
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