<s>[INST] Find the extreme values of the given function, subject to constraints $f(x,y) = y^2 - x^2 ; x^2 + y^2 = 36$ 
Find the extreme values of the function, subject to constraints
$$f(x,y) = y^2 - x^2 ; x^2 + y^2 = 36$$

Here is what I have done:
Using the substitution method we can say:
$$y = \sqrt{-x^2 + 36}$$
Then plugging that into our original equation:
$$f(x,y) = (\sqrt{-x^2 + 36})^2 - x^2$$ which simplifies to $$-2x^2+36$$
now $$f_x(x,y) = 2(-2x) + 0$$
set that $=0$ and get $x=0$
Now find if min/max:
$$f_{xx} = -4 $$
$$-4 < 0; maximum$$
Answer: There is a maximum point at $x=0$?
Is this right?
 [/INST] Let $x = 6\cos \theta, y = 6\sin \theta \implies F(x,y) = F(\theta) = 36\cos (2\theta)\le 36$, and the max is $36$. The min is $-36$ and this occurs when $\cos (2\theta) = -1$ which corresponds to $x^2 - y^2 = -36$. Since $x^2+y^2 = 36$, we have: $x = 0, y = \pm 6$.
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