Calculus

Lagrange Multipliers

Stand the objective up as a hill and the constraint becomes a fence winding across it. Walk the fence to its highest point: there the ground is tangent to a contour line, which is exactly the statement that the two gradients line up.

the fence, f along the constraint∇f∇gthe ball, a peak on the fence

Drag to orbit · scroll to zoom · the dashed line is the contour through the ball · Walk uphill climbs the fence until the slope along it is zero

Slope along the fence—

—

Objective f(x, y)

Where on the fence 40°

Circle radius r (g = x² + y² = r²) 1.00

Show the constraint wallSpin the scene

Values

height f here—

∠(∇f, ∇g)—

multiplier λ here—

best height f* on this circle—

df* / d(r²)—

What to observe

Press Walk uphill: the ball climbs the fence and stops where the slope along it is zero. There the dashed contour through the ball just touches the fence.
At the top ∇f and ∇g are 0° apart, so ∇f = λ∇g. Away from the peaks and dips they disagree and there is still uphill left.
Change the radius r and the best height changes with it. At the top, λ equals df*/d(r²): the multiplier is the price of loosening the constraint.
On x · y the fence has two peaks and two dips. Walking uphill finds the peak on your side, not always the same one.

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