For positive x and y with x + y = 10, find the maximum of x^2 y^3 and where it occurs.

For positive x and y with x + y = 10, find the maximum of x^2 y^3 and where it occurs.

Approach: Split the constraint into five equal pieces matched to the exponents so the arithmetic mean geometric mean inequality applies with all terms equal at the optimum.

3456. Write x + y = x/2 + x/2 + y/3 + y/3 + y/3, a weighted splitting into five terms matched to the exponents 2 and 3. The arithmetic mean geometric mean inequality gives (x/2)^2 (y/3)^3 <= ((x + y)/5)^5 = 2^5 = 32, so x^2 y^3 <= 32 * 4 * 27 = 3456. The equality condition is that all five terms are equal, so x/2 = y/3, which with x + y = 10 gives x = 4 and y = 6, and 4^2 * 6^3 = 16 * 216 = 3456 confirms the constrained maximum. Calculus reaches the same point by maximising 2 log x + 3 log y with a Lagrange multiplier, giving 2/x = 3/y.

Follow-up: Maximise x^a y^b subject to x + y = S for positive a and b, and say how the answer degenerates as b grows without bound?

Key concepts: am-gm inequality, equality condition, constrained maximum, weighted splitting.