Question

Use lagrange multipliers to find the points on the given surface that are closest to the origin. y2 = 64 + xz

Answer 1

The distance between an arbitrary point on the surface and the origin is

$d(x,y,z)=\sqrt{x^2+y^2+z^2}$

Recall that for differentiable functions $g(x)$ and $h(x)$, the composition $g(h(x))$ attains extrema at the same points that $h(x)$ does, so we can consider an augmented distance function

$D(x,y,z)=x^2+y^2+z^2$

The Lagrangian would then be

$L(x,y,z,\lambda)=x^2+y^2+z^2+\lambda(y^2-64-xz)$

We have partial derivatives

$\begin{cases}L_x=2x-\lambda z\\L_y=2y+2y\lambda\\L_z=2z-\lambda x\\L_\lambda=y^2-64-xz\end{cases}$

Set each partial derivative to 0 and solve the system to find the critical points.

From the second equation it follows that either $y=0$ or $\lambda=-1$. In the first case we arrive at a contradiction (I'll leave establishing that to you). If $\lambda=-1$, then we have

$\begin{cases}2x+z=0\\2z+x=0\end{cases}\implies x=0,z=0$

This means $y^2=64\implies y=\pm8$

so that the points on the surface closest to the origin are $(0,\pm8,0)$.