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Minimum of x²+xy+y²+3z² under constraint

Short answer

Solve ∇f = k∇g for f = x²+xy+y²+3z² and constraint x+2y+4z=60. Stationary point: x=0, y=90/7, z=60/7. The positive definite form ensures a minimum: 2700/7.

Find the minimum value of the function x²+xy+y²+3z² under the condition x+2y+4z = 60.

Minimise f = x² + xy + y² + 3z² subject to g = x + 2y + 4z − 60 = 0.

Lagrange conditions

Let F = f + λ(x + 2y + 4z − 60). Then (taking −λ as the multiplier k, so ∇f = k∇g)

2x + y  = k          ... (1)
x + 2y  = 2k         ... (2)
6z      = 4k         ... (3)
x + 2y + 4z = 60     ... (4)

Solving

  • (2) − 2·(1): x + 2y − 4x − 2y = 2k − 2k, so −3x = 0, hence x = 0.
  • Then (1) gives y = k, and (2) gives 2y = 2k, which is consistent.
  • From (3), z = 2k/3.
  • Put these in (4): 0 + 2k + 8k/3 = 60, so 14k/3 = 60, k = 90/7.

So x = 0, y = 90/7, z = 60/7.

Value

f = y² + 3z² = (90/7)² + 3(60/7)²
  = 8100/49 + 10800/49
  = 18900/49 = 2700/7 ≈ 385.71

Why it is a minimum

The quadratic form x² + xy + y² + 3z² has a matrix with eigenvalues 1/2, 3/2 and 3 (all positive), so f is positive definite and convex. A convex function restricted to a plane has its minimum at its stationary point.

Answer: minimum value = 2700/7 ≈ 385.71, at (0, 90/7, 60/7).

✓ Answer checked by Claude Opus 5.5 (high effort)

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