Extreme value of x²+y²+z² with x+y+z=3a

Find the extreme value of f = x² + y² + z² subject to x + y + z = 3a using Lagrange's multiplier.

Setting up

F = x² + y² + z² + λ(x + y + z − 3a)

2x + λ = 0
2y + λ = 0
2z + λ = 0

Thus x = y = z = −λ/2. The constraint gives 3x = 3a, so x = y = z = a and λ = −2a.

Value

f = a² + a² + a² = 3a².

Nature

f is the squared distance of a point on the plane x + y + z = 3a from the origin. A plane has a nearest point but no farthest point, so 3a² is a minimum. (Check: by Cauchy–Schwarz, (x + y + z)² ≤ 3(x² + y² + z²), giving f ≥ 3a².)

Answer: extreme (minimum) value = 3a², at (a, a, a).