Finding maxima and minima of $f(x,y)= x^4 + y^4 - 2x^2 - 2y^2 + 4xy$

Question

For $f(x,y)=x^4+y^4-2x^2-2y^2+4xy$, I need to find maxima or minima. There are three critical points: $(0,0),(\sqrt2, -\sqrt2),(-\sqrt2,\sqrt2)$ So at $(\sqrt2, -\sqrt2)$, $f$ has minimum value, $-8$ and at $(-\sqrt2,\sqrt2),$ it has same minimum value, $-8$. At $(0,0)$, after inspecting, I get that it has neither maxima or minima. But when we substitute $(0,0),$ we get $0\le f \le8,$ so should not we get $(0,0)$ as a point of maxima?

Answer 1

More generally you can just use the Hessian at the critical points. We have $$ \begin{aligned} f_{xx}&=12x^2-4,\\ f_{xy}&=4,\\ f_{yy}&=12y^2-4. \end{aligned} $$ So at the critical points $P_{1,2}=\pm(\sqrt2,\sqrt2)$ we have $$H_f(P_{1,2})=\left(\begin{array}{cc}12\cdot2-4&4\\4&12\cdot2-4\end{array}\right).$$ The principal minors $\Delta_1=20$, $\Delta_2=20^2-4^2=384$ are thus both positive. Meaning that the Hessian form is positive definite, and we have local minima.

On the other hand at the remaining critical point $P_3=(0,0)$ the Hessian $$H_f(P_3)=\left(\begin{array}{cc}12\cdot0-4&4\\4&12\cdot0-4\end{array}\right)$$ is negative semidefinite as the determinant $=\Delta_2=0$.

This is immediately obvious from the quadratic Taylor polynomial at $P_3$ as well, because $$ T_{2,P_3}(x,y)=-2x^2-2y^2+4xy=-2(x-y)^2. $$ The nature of this critical point is thus left in the dark by a study of the Hessian alone. Let's do some testing suggested by the form of the quadratic Taylor polynomial. If $x=y=h$, then $$ f(x,y)=f(h,h)=2h^4-2(h-h)^2=2h^4>0. $$ On the other hand $$ f(h,-h)=2h^4-2(h-(-h))^2=-8h^2+2h^4<0 $$ when $h$ is sufficiently close to zero.

This means that $P_3$ is not a local extremal point.

Michael Rozenberg already handled the global minima/maxima.

Answer 2

It's obvious that a maximal value does not exist.

We'll prove that $-8$ is a minimal value.

Let $x=\sqrt2a$ and $y=-\sqrt2b$.

Hence, we need to prove that $$4a^4+4b^4-4a^2-4b^2-8ab+8\geq0$$ or $$a^4+b^4+2\geq(a+b)^2,$$ which is AM-GM and C-S: $$a^4+b^4+2\geq2\sqrt{(a^4+1)(1+b^4)}\geq2(a^2+b^2)\geq(a+b)^2$$