Consider the surface $\Sigma:=\{(x,y,z)\in\mathbb{R}^3:xy-z^2=16\}$. I want to find the point on $\Sigma$ nearest to origin but I have troubles whit this.
I define the function $f(x,y)=\sqrt{xy-16}$ and I tried find the minimun of the function $g(x,y)=d^2(x,y)=x^2+y^2+xy-16$ (Here the function $d=||(x,y,z)||$). My problem is that the only critic point of $g(x,y)$ is in origin but this point in not in $\Sigma.$
Can someone give me a hint to solve my problem?
Use the method of Lagrange multipliers to minimize the function $D(x,y,z) = x^2 + y^2 + z^2$ subject to the constraint $g(x,y) = xy - z^2 - 16 = 0$.
You are looking at solving the system $\nabla D = \lambda \nabla g$, $g = 0$, i.e. \begin{align*} 2x &= \lambda y \\ 2y &= \lambda x \\ 2z &= \lambda(-2z) \\ xy - z^2 - 16 &= 0.\end{align*}
The third equation gives you either $z = 0$ or $\lambda = -1$. Work from there.