If x and y are positive real numbers such that x²y³ = 32, then the least value…

Answer: 10

Use Lagrange multipliers to minimize f(x,y) = 2x + 3y subject to the constraint x² y³ = 32.

1. Set g(x,y) = x² y³ - 32. Solve ∇f = λ ∇g: 2 = λ·(2x y³) and 3 = λ·(3 x² y²).

2. From the two equations, λ = 1/(x y³) and λ = 1/(x² y²). Equating gives x y³ = x² y², and since x,y > 0 this simplifies to y = x.

3. Substitute y = x into the constraint: x²·x³ = x^5 = 32, so x = 32^(1/5) = 2. Hence y = 2.

4. Compute 2x + 3y = 2·2 + 3·2 = 4 + 6 = 10. Therefore the least value is 10, attained at x = y = 2.

Alternative approach: a weighted AM-GM argument leads to the same conclusion that equality occurs when x = y, giving x^5 = 32 and the same minimal value.