Lagrange Multipliers

• Extrema of $f(\vec{r})$ subject to constraint $g(\vec{r})=c$ satisfy $(\nabla f=\lambda \nabla g\wedge g=c)$
• Extrema of $f(\vec{r})$ subject to constraints $g(\vec{r})=c_1,h(\vec{r})=c_2$ satisfy $(\nabla f=\lambda \nabla g+\mu \nabla h\wedge g=c_1,h=c_2)$
• Extrema $⟹\nabla f\propto \nabla g$ is a consequence of:
  • The Gradient of $g$ being normal to the level curve $g=c$, everywhere
  • The gradient of $f$ along the boundary imposed by the constraint is also normal to the level curve $g=c$, wherever the Parameterization of $f$ along the boundary ($f(t)$; single-variable calculus) would have a zero derivative, i.e. where there are possible Extrema. $f$ along the boundary is normal to the level curve at possible extrema because its parameterization is flat.
• See here for a great video explanation

Examples

Example 1

Find the points on surface $z^2=xy+4$ that are closest to the origin. Let $d(x,y,z)=\sqrt{x^2+y^2+z^2}$ be the distance from a point to the origin Let $f(x,y,z)=x^2+y^2+z^2$ be the proxy to the distance function that we will use to solve this problem, because its extrema exactly correspond to that of $d$. $f$ is our objective function Let $g(x,y,z)=z^2-xy=4$ be our constraint function Goal: find points where $\nabla f\propto \nabla g$ and $g=c$ Find Gradients: $\nabla f=<2x,2y,2z>\nabla g=<-y,-x,2z>$ Solve this system:

1. $2x=-\lambda y$
2. $2y=-\lambda x$
3. $2z=\lambda 2z$
4. $z^2-xy=4$ Blah blah blah… You do a bunch of different cases and get the following points: $(2,-2,0),(-2,2,0),(0,0,2),(0,0,-2)$ You can plug each of these into $f$ and you’ll find $\max f=8⟹\max d=\sqrt{2}$ and $\min f=4⟹\min d=2$