This figure demonstrates the method of Lagrange mulipliers to solve a problem like the following:[br][br][b]Suppose functions [math]f(x,y)[/math] and [math]g(x,y)[/math] are given. [/b][b]Find the local maximum and minimum values of [math]f[/math] subject to the constraint that [math]g(x,y) = 0[/math].[/b][br][br]Graphically, the equation [math]g(x,y) = 0[/math] defines a curve in the plane, call it [math]C[/math]; we must find the point(s) on [math]C[/math]where [math]f[/math] reaches its extreme values. [br][br]When the figure loads, [math]f[/math] and [math]g[/math] are defined and you see curve [math]C[/math], a point on [math]C[/math], and the vector [math]\nabla f[/math] at the point. Keep in mind that [math]\nabla f[/math] points in the direction of greatest increase of [math]f[/math]. With this in mind, slowly drag the point around the curve - can you get a sense of where [math]f[/math] is increasing or decreasing along the curve? What makes [math]f[/math] attain a maximum or minimum value along the curve?[br][br]When [math]f[/math] reaches an extreme value on [math]C[/math], the gradient [math]\nabla f[/math] must be perpendicular to [math]C[/math] - this is the Orthogonal Gradient Theorem. Since [math]\nabla g[/math] is always orthogonal to [math]C[/math], this is equivalent to the equation [math]\nabla f=\lambda\nabla g[/math] for some scalar [math]\lambda[/math]. By finding the points on [math]C[/math] where the equation is satisfied we can determine where [math]f[/math] takes on its extreme values.[br][br]Finally, in the figure you can draw level curves for [math]f[/math] and explore the relationship between level curves of [math]f[/math], the curve [math]C[/math], and the location of the extreme points.

[i]Developed for use with Thomas' Calculus and [url=https://www.pearson.com/en-us/subject-catalog/p/interactive-calculus-early-transcendentals-single-variable/P200000009666]Interactive Calculus[/url], published by Pearson.[/i]