Maximum of xyz with x+y+z=24

Maximise f = xyz subject to g = x + y + z − 24 = 0 using Lagrange's multipliers (x, y, z > 0).

Setting up

Let F = xyz + λ(x + y + z − 24). Stationary conditions:

∂F/∂x = yz + λ = 0
∂F/∂y = xz + λ = 0
∂F/∂z = xy + λ = 0
x + y + z = 24

Solving

From the first three equations, yz = xz = xy = −λ.

• yz = xz gives z(y − x) = 0, so x = y (z ≠ 0).
• xz = xy gives x(z − y) = 0, so y = z.

Hence x = y = z. Using the constraint, 3x = 24, so x = y = z = 8 (and λ = −64).

Nature of the point

The constraint set with x, y, z > 0 is a bounded triangle, and xyz → 0 on its boundary while xyz > 0 inside, so the single stationary point must be a maximum. (Check with AM–GM: (xyz)^(1/3) ≤ (x+y+z)/3 = 8, so xyz ≤ 512.)

Answer: maximum value of xyz = 8 × 8 × 8 = 512, at x = y = z = 8.