How to solve for $x$ and $y$ given these two multivariable functions?

Question

I am trying to find the critical points of this function $(x-y)(16-xy)$

I found that the partial derivatives are $f_x = 16-2xy+y^2 = 0$ and $f_y=-x^2-16+2xy = 0$. However, I have tried everything but cannot solve for $x$ and $y$ with these two equations. How do you solve for $x$ and $y$?

Answer 1

If you solve $f_x = 0$ for e.g. x you get $x = \frac{16 + y^2}{2 y}$. Insert this into $f_y$ you get to: $-8 - \frac{64}{y^2} + \frac{3 y^2}{4} = 0$ You can solve this e.g. with multiplying with $y^2$, substituting $y^2$ then and get two real solutions: $y_1 = -4$, $y_2 = 4$. Inserting this back into your $f_x$ solution you get the corresponding $x$ solutions. Assuming you are just looking for real solutions.

Answer 2

Note $$2xy-y^2=16=2xy-x^2$$ which leads to $x^2=y^2$, or $x=\pm y$.