Maximum and minimum value of a function.

Question

Let us consider a function $f(x,y)=4x^2-xy+4y^2+x^3y+xy^3-4$. Then find the maximum and minimum value of $f$.

My attempt. $f_x=0$ implies $8x-y+3x^2y+y^3=0$ and $f_y=0$ implies $-x+8y+x^3+3xy²=0$ and $f_{xy}=3x^2+3y^2-1$. Now $f_x+f_y=0$ implies $(x+y)((x+y)^2+7)=0$ implies $x=-y$ as $x$ and $y$ are reals. Now putting it in $f_x=0$ get the three values of $x$ as $0$, $(-3+3\sqrt{5})/2$ and $(-3-3\sqrt{5})/2$. And then $f_{xy}(0,0)<0$ implies $f$ has maximum at $(0,0)$. But at other two points $f_{xy}$ gives the positive value. So how can I proceed to solve the problem. Please help me to solve it.

Answer 1

Note that $f(x,y)=(xy+4)(x^2+y^2-1)$ and therefore $$\lim_{x\to+\infty}f(x,x)=\lim_{x\to+\infty}(x^2+4)(2x^2-1)=+\infty$$ and $$\lim_{x\to+\infty}f(x,-x)=\lim_{x\to+\infty}(-x^2+4)(2x^2-1)=-\infty.$$ What may we conclude?

P.S. The critical point $(0,0)$ is a local minimum.

Answer 2

$f_x+f_y=\left(x^3+3 x y^2-x+8 y\right)+\left(3 x^2 y+8 x+y^3-y\right)=\\=x^3+3 x^2 y+3 x y^2+7 x+y^3+7 y=(x+y) \left(x^2+2 x y+y^2+7\right)=0$

$y=-x$. Substitute in $f_x=0$ and get

$9 x-4 x^3=0\to x_1=0;\;x_2=\dfrac{3}{2};\;x_3=-\dfrac{3}{2}$

and $y_1=0;\;y_2=-\dfrac{3}{2};\;y_3=\dfrac{3}{2}$

Now the Hessian, for the second partial derivative test

$H(x,y)=\left( \begin{array}{rr} 6 x y+8 & 3 x^2+3 y^2-1 \\ 3 x^2+3 y^2-1 & 6 x y+8 \\ \end{array} \right)$

and $\det H(x,y)=-9 x^4+18 x^2 y^2+6 x^2+96 x y-9 y^4+6 y^2+63$

so we have

$\det H(x_1,y_1)=63>0$ and $f_{xx}=8>0$ thus $(0,0)$ is a local minimum

$\det H(x_2,y_2)=\det H(x_3,y_3)=-126<0$ so $\left(\frac{3}{2}, -\frac{3}{2}\right)$ and $\left(-\frac{3}{2}, \frac{3}{2}\right)$ are saddle points

Hope this helps

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