Find the point on curve nearest to the origin

Question

The curve is $xy^2=54$ and the function $f(x,y)=x^2+y^2$ whose minimum value I want to compute on curve. Now I want to find the minimum value of function subject to the constrain that is $g(x,y)=xy^2-54=0$. Using the method of Lagrange multiplier i.e, $\vec{\nabla}f=\lambda \vec{\nabla}g$ and $g(x,y)=0$. I get $2x=\lambda y^2$, $2y=2\lambda xy$ and $xy^2-54=0.$ I tried a lot to find to find point on the curve from these three equation but I am facing troubles. Two variables can't be eleminate at time from these three. THanks in advance!

Answer 1

f(x,y) = x$^2$ + 54/x = g(x).
g'(x) = 2x - 54/x$^2$ = 0.
2x$^3$ = 54, etc.

Answer 2

You can get the solution by eliminating one of the variables using the constraint, as said in other answer. But you may prefer follow the method you chose to the end.

Three equations can be enough to find three unknowns,

$$\left\{ \begin{matrix} 2x=\lambda y^2\\ 2y=2\lambda xy \\ xy^2-54=0\\ \end{matrix} \right.$$

or

$$\left\{ \begin{matrix} 2x=\lambda y^2\\ y(\lambda x-1)=0 \\ xy^2-54=0\\ \end{matrix} \right.$$

From the second equation have that $y=0$, leading to the contradiction $0=54$ in the third one, or $\lambda=1/x$. Plug the value of $\lambda$ into the first: $2x^2=y^2$ and now, the value for $y^2$ found into the third: $2x^3-54=0$. An equation with a single real solution, $x=3$. Then $y=\pm3\sqrt 2$

The two solutions produce the same value for $f$ and it itself is not bounded, thus the solutions are minima.