Extreme value of x²+y²+z² with ax+by+cz=p

Find the stationary value of f = x² + y² + z² subject to g = ax + by + cz − p = 0, using Lagrange's multiplier λ.

Setting up

F = x² + y² + z² + λ(ax + by + cz − p)

∂F/∂x = 2x + λa = 0   →  x = −λa/2
∂F/∂y = 2y + λb = 0   →  y = −λb/2
∂F/∂z = 2z + λc = 0   →  z = −λc/2

Finding λ

Put these into the constraint ax + by + cz = p:

−(λ/2)(a² + b² + c²) = p, so λ = −2p/(a² + b² + c²).

Then

x = ap/(a² + b² + c²)
y = bp/(a² + b² + c²)
z = cp/(a² + b² + c²)

Extreme value

x² + y² + z² = p²(a² + b² + c²)/(a² + b² + c²)²
             = p²/(a² + b² + c²)

Nature

x² + y² + z² is the squared distance of a point of the plane ax + by + cz = p from the origin. A plane has a nearest point but no farthest point, so the value is a minimum.

Answer: extreme (minimum) value = p²/(a² + b² + c²).