Maximizing f(x,y) = xy subject to x² + y² = 8

Calculus 3 · Axiom Academy

EXAMPLE Maximizing f(x,y)=xy subject to x+y=8 Using Lagrange multipliers to maximize a product on a linear constraint Find the maximum value of f(x,y) = xy subject to the constraint g(x,y) = x + y = 8 , using the method of Lagrange multipliers. At the maximum, the constraint line x+y=8 just touches the level curve xy=16 at (4,4) — the two curves are tangent, so their gradients are parallel. Smaller products (like xy=8 ) cross the line; larger ones never reach it. Nice work — you maximized a product on a linear constraint with Lagrange multipliers. Here's what carried the solution: Set up ∇f = λ∇g: With and , the equations are and . Eliminate λ: Both partials equal , so x = y immediately. Apply the constraint: Substituting into x + y = 8 gives 2x = 8 , so x = y = 4 and . Geometric insight: At the maximum the line x+y=8 is tangent to the level curve xy=16 ; parallel gradients are exactly what demands. Quick sanity check: on the line, f = x(8-x) = 8x - x^2 , so f'(x) = 8 - 2x = 0 gives x = 4 and f = 16 — the same answer without multipliers.

This is the written version of the interactive lesson above. See the full Calculus 3 course.