Lagrange Multiplier Calculator

The Lagrange multiplier method finds extrema of a function f(x,y) subject to a constraint g(x,y) = 0. The key insight: at an extremum on the constraint curve, the gradient of f must be parallel to the gradient of g — they cannot point in independent directions, or you could move along the constraint to increase f.

So the method solves: ∇f = λ ∇g, plus the constraint g = 0. The scalar λ (the Lagrange multiplier) is auxiliary and rarely matters by itself.

This calculator handles two standard constraint shapes with a linear objective f(x,y) = ax + by.

Circle constraint x² + y² = r²: ∇f = (a, b), ∇g = (2x, 2y). Setting (a,b) = λ(2x, 2y) gives x = a/(2λ), y = b/(2λ). Substituting into the constraint: (a² + b²)/(4λ²) = r², so 2λ = ±√(a² + b²)/r.

Critical points: ±r · (a, b) / √(a² + b²). The maximum is f_max = +r · √(a² + b²); the minimum is the negative.

Ellipse constraint x²/p² + y²/q² = 1: The same setup but with weighted gradient ∇g = (2x/p², 2y/q²). Critical points have x = ap²/(2λ), y = bq²/(2λ). Constraint substitution gives 2λ = ±√(a²p² + b²q²). Critical points: ±(ap², bq²)/√(a²p² + b²q²). The extremum values are ±√(a²p² + b²q²).

Worked example: maximize 3x + 4y on the circle of radius 5. f_max = 5 × √(9+16) = 5 × 5 = 25, achieved at (3, 4). Minimum is -25 at (-3, -4). Sanity check: the gradient (3,4) is exactly the position vector at the maximum point, so they are parallel as required.

The geometric picture: the level curves of f = ax + by are parallel lines. The maximum on the constraint curve is the line farthest in the gradient direction that still touches the curve — that touchpoint is exactly where ∇f and ∇g are parallel.

Lagrange multipliers extend to multiple constraints (one λ per constraint) and to nonlinear objectives. The general principle remains: at any constrained extremum, the gradient of f lies in the span of the constraint gradients.