Find the extreme values of the function

Question

Find the extreme values of the following expression. $$x^4 + y^4 - 2x^2 + 4xy - 2y^2. $$

I have tried using the partial derivatives method for finding critical points but that's giving me 2 equations (partially differentiating with respect to x and y) which look pretty unsolvable. Any other methods (college level) are more than welcome. Thank you.

Answer 1

Your function is $f(x,y)=x^4+y^4-2(x-y)^2$.

From the two partial derivatives we find the system $$ \begin{cases} x^3-(x-y)=0\\ y^3+(x-y)=0 \end{cases} $$ adding the two equation this gives $x=-y$ and substituting in the first equation we have:

$x^3-2x=0$ that gives the solutions $x=0$ and $x=\pm \sqrt{2}$.
Can you do from this?

Answer 2

For $y=0$ and $x\rightarrow+\infty$ we see that $x^4+y^4-2x^2+4xy-2y^2\rightarrow+\infty$,

which says that the maximal value does not exist.

For $x=\sqrt2$ and $y=-\sqrt2$ we get a value $8$.

We'll prove that it's a minimal value.

Indeed, we need to prove that $$x^4+y^4-2x^2+4xy-2y^2\geq-8$$ or $$x^4+y^4+8\geq2(x-y)^2.$$

Now, by AM-GM $$x^4+y^4+8\geq2\sqrt{8(x^4+y^4)}.$$ thus, it remains to prove that $$\sqrt{8(x^4+y^4)}\geq(x-y)^2.$$ Let $x^2+y^2=2kxy$.

Thus, $|k|\geq1$ and we need to prove that $$8((x^2+y^2)^2-2x^2y^2)\geq(x^2+y^2-2xy)^2$$ or $$8(4k^2-2)\geq(2k-2)^2$$ or $$(k+1)(7k-5)\geq0,$$ which is obviously true for $|k|\geq1$.

Done!