Find the points on the surface $xy^2z^4=14$ that are closest to the origin.

Question

Find the points on the surface $x y^2 z^4=14$ that are closest to the origin.

this was my exam question. Our teacher said she wont publish answers. I couldn't even move my pen with this question.

Answer 1

Hint: The function which you have to minimize is $d(x,y,z)=\sqrt{x^2+y^2+z^2}$, but there is a side condition (or a constraint): $xy^2z^4=14 \iff xy^2z^4-14=0 \quad (\varphi(x,y,z)=0)$.

Hence, compose Lagrange function $F(x,y,z,\lambda)=d(x,y,z)-\lambda\varphi(x,y,z)$, the extreme value of $F$ will be an extreme value also for $d$.

Answer 2

The distance is

\begin{align} & (x^2+y^2+z^2)^{1/2}\\ =& \left(x^2+2(\frac12y^2)+4(\frac14z^2)\right)^{1/2}\\ \ge &7 \left( \frac{x^2y^4z^8}{2^2\cdot 4^4}\right)^{1/{14} } = \frac{7^{9/14}}{2^{4/7} } \end{align} where the AM-GM inequality is applied and the equality, or the minimum distance, occurs at $x^2=\frac{y^2}2= \frac{z^2}4$, which leads to the closest points $x=(\frac7{16})^{1/7}$, $y= \pm \sqrt2(\frac7{16})^{1/7} $ and $z= \pm 2(\frac7{16})^{1/7} $.