There is a problem on WebWork that a few of you are asking about. Problem 18 in the Section 15.1 & 15.2 assignment asks the following: What is the shortest distance from the surface $xy+3x+z^2=12$ to the origin?

**Solution:** When you’re dealing with a distance problem, you’re tempted to minimize something that looks like

\[f(x,y,z)=\sqrt{x^2+y^2+z^2}.\]

I’d caution against doing that, since you might as well minimize $f(x,y,z)=x^2+y^2+z^2$ instead (it’s a LOT easier). We’ll set $S=x^2+y^2+z^2$ and we’ll take the given function and solve for $z$ as best as possible, getting $z^2=12-xy-3x$. Putting these two pieces together, we end up with

\[S=x^2+y^2+12-xy-3x.\]

This is a polynomial function, so we know that the critical points will only occur where the gradient is zero. So, we have

\[\nabla S=\langle 2x-y-3,2y-x\rangle =\vec{0}.\]

This implies that $x=2y$, and it follows that $(2,1)$ is the only critical point. Now, we use the second derivative test, where

\[D=S_{xx}(2,1)S_{yy}(2,1)-(S_{xy}(2,1))^2=3>0\]

and so we know that we’re at a local minimum. Moreover, looking at $S$ as $x\to\pm\infty$ and $y\to\pm\infty$, we have $S$ going to positive infinity in all directions. Thus, we know that we have found the global minimum of our distance function.

## no comment untill now