An example of Lagrange Multiplier Method to illiustrate Theorem 9 of the expository article by Prof. S. Kumaresan on Implicit Function Theorem (finite dimensional version). The article can be found here {http://accounts.mtts.org.in/download-article.php?articleid=79}.

Theorem: (Lagrange Multiplier)[br][br]Let [math]f:U\subset\mathbb{R}^n\longrightarrow\mathbb{R}^n[/math] and [math]g:U\subset\mathbb{R}^n\longrightarrow\mathbb{R}^n_{ }[/math] be [math]C^1[/math]. Let [math]p\in U,[/math] [math]g\left(p\right)=\alpha[/math] and [math]S:=g^{-1}\left(\alpha\right),[/math] the level set of [math]g[/math] at [math]\alpha[/math]. Assume that [math]\nabla g\left(p\right)\ne0.[/math] If the restriction of [math]f[/math] to [math]S[/math] has a local extremum at p, then there is a real number [math]\lambda[/math] such that [math]\nabla f\left(p\right)=\lambda\nabla g\left(p\right).[/math][br][br]Example: Find the extrema of [math]f\left(x,y\right)=xy[/math] subject to the constraint [math]\frac{x^2}{9}+\frac{y^2}{4}=1.[/math][br][br]It is easy to find [math]\lambda[/math] by solving [math]\nabla f\left(x,y\right)=\lambda\nabla g\left(x,y\right)[/math]. On eliminating [math]\lambda[/math] from [math]x[/math] and [math]y[/math] we can find the points on the ellipse at which [math]f[/math] takes extreme values. The answer is [math]\lambda=\pm3,[/math] and the points of extrema are [math]\left(\pm\frac{3}{\sqrt{2}},\pm\sqrt{2}\right).[/math] [br][br]The points that lie on ellipse [math]g[/math] forms the feasible set. The purple lines are level curves of the graph of [math]f[/math]. The slider [math]h[/math] is the value of the objective function [math]f[/math] at the green dots on the feasible set. Note that as the value of [math]\left|h\right|>3[/math] the green dots disappear from the picture showing that we are out of the feasible set.[br]The two black dots are the points of maximum and the maximum value is [math]3[/math]. The points of intersection of ellipse with the level curve (dotted red line) at [math]h=-3[/math] are the points of minimum. Note that there is a common tangent at the point of extrema to the ellipse and the level curves. The normals at one of the maximum to the constraint and objective function are marked by a ray with respective colors.[br][br]Hint to use this:[br][br]Use the slider(on the left) to move the horizontal plane(red shaded) parallel to the xy plane. As you move the slider, the corresponding level curve (dotted red colored hyperbola) moves in the left side view. The level curve is the intersection of the red plane with the green surface (graph of [math]f[/math]) .[br][br]Apart from this the right side view can be rotated in all possible directions to see the complete picture.