finding points on curve $x^2y^2=16$ that are closest to origin in two ways

Question

(a) by finding the minimum value of an appropriate function of one variable

(b) by using the method of Lagrange multipliers.

so from what I understand, for part (a) I can use the equation that $x^2y^2=16$ and write $y$ in terms of $x$ in the equation $x^2+y^2$. After that, I will find the gradient to find critical points and use the Hessian matrix to determine the type of extremum.

And for part (b), let $f(x,y)=x^2+y^2$ and $g(x,y)=x^2y^2$ and then

$\nabla f=\lambda \nabla g$ and solve the equations.

Am I on the right path?

Edit:- So I solved the answer through the first method and I have found the critical points through Lagrange multipliers method. But how do I write the Hessian matrix to prove that those critical points are minimal? I am having trouble understanding it. I only know that the first element should be 0 and other elements of the first row and column should be $[-2xy^2\ -2x^2y]$. I can't proceed from there.

Answer 1

For (b) you are on the right path.

For (a): we have $y^2=\frac{16}{x^2}$, hence $x^2+y^2=x^2+\frac{16}{x^2}$.

Therefore you have to minimize the function $g(x):= x^2+\frac{16}{x^2}$.

Answer 2

An option:

Want to minimize $x^2+y^2$:

(Note : This is the distance$^2$, easier to handle).

Let $x,y >0$, i.e. the first quadrant.

$x^2+y^2 = (x-y)^2 +2xy =(x-y)^2 + (2)4.$

Minimum : $x=y$.

$2x^2= (2)(4)$, or $x=2$, and $y=2$.

Considering the invariance of the problem under

$x \rightarrow -x$, and $y \rightarrow -y$ ,

you get the remaining $(x,y)$ coordinates.