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Extreme value with two linear constraints

Short answer

Use Lagrange multipliers with two constraints. Solve for x, y, z from partial derivatives and constraints, obtaining x=1/3, y=2/3, z=2/3. The function is squared distance from origin to a line, so only a minimum exists. Minimum value is 1.

Find the extreme value of x²+y²+z² connected by the relation x+z = 1 and 2y+z = 2.

Find the extreme value of f = x² + y² + z² subject to g₁ = x + z − 1 = 0 and g₂ = 2y + z − 2 = 0.

Lagrange conditions

F = x² + y² + z² + λ(x + z − 1) + μ(2y + z − 2)

∂F/∂x = 2x + λ = 0        →  x = −λ/2
∂F/∂y = 2y + 2μ = 0       →  y = −μ
∂F/∂z = 2z + λ + μ = 0    →  z = −(λ + μ)/2

Using the constraints

Let a = −λ/2 and b = −μ. Then x = a, y = b, z = a + b/2.

  • x + z = 1: a + a + b/2 = 1, so 2a + b/2 = 1.
  • 2y + z = 2: 2b + a + b/2 = 2, so a + 5b/2 = 2.

Solve: from the first, b = 2 − 4a. Put into the second: a + 5 − 10a = 2, so a = 1/3, b = 2/3.

So x = 1/3, y = 2/3, z = 1/3 + 1/3 = 2/3.

Check: x + z = 1 ✓. 2y + z = 4/3 + 2/3 = 2 ✓.

Value

f = 1/9 + 4/9 + 4/9 = 1.

Nature

The two constraints define a straight line, and f is the squared distance from the origin to a point of the line. A line has a nearest point but no farthest, so the value is a minimum.

Answer: minimum value = 1, at (1/3, 2/3, 2/3).

✓ Answer checked by Claude Opus 5.5 (high effort)

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