Method of Lagrange Multipliers

Calculus 3 · Axiom Academy

LESSON Method of Lagrange Multipliers A powerful technique for finding extrema subject to constraints — built on the elegant geometry of parallel gradients. 1. The Constrained Optimization Problem We want to optimize — find the maximum or minimum of — a function f(x,y) subject to a constraint g(x,y)=c . Without the constraint we could just hunt for critical points where ; the constraint forces us to search only along the curve g(x,y)=c . Maximize the area of a rectangle with fixed perimeter 40 . Objective: f(x,y)=xy (area) · Constraint: g(x,y)=2x+2y=40 (perimeter). At a constrained extremum, a level curve of f just touches the constraint curve — they are tangent there. Because both curves share that tangent line, the gradients and (each perpendicular to its own curve) must point along the same line: they are parallel . Since the gradients are parallel, we can write for some scalar , called the Lagrange multiplier . (For the rectangle, at (10,10) we get , so .) The parallel-gradient condition turns the geometry into algebra. To find constrained extrema we solve the system: Find and in terms of x and y . Write the equations from together with g=c . Solve for — usually by eliminating first. Plug each solution into f to see which gives the max or min. For our rectangle: forces x=y ; combined with 2x+2y=40 this gives the critical point below.

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