# Finding critical points and determining local maxima and minima for $f(x,y)=-x^3+4xy-2y^2+1$

Finding critical points and determining local maxima and minima for $f(x,y)=-x^3+4xy-2y^2+1$

First I took the following derivatives:

$f_x=-3x^2+4y$

$f_{xx}=-6x$

$f_y=4x-4y$

$f_{yy}=-4$

$f_{xy}=4$

I then set $f_x=0$ and $f_y=0$ and solved for the solution to the system of equations. The answer I got is $x=\frac{4}{3}$ and $y=\frac{4}{3}$

I know I now have to classify my critical points to determine if they are local maxima or minima. So using the following formula:

$D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-[f_{xy}(x,y)]^2$

I get:

$-6x(-4)-(4)^2 = 24x-16$

For the next step I'm supposed to plug in $D(x,y)$ but I have no $y$ variable. Any hints on where I'm going wrong here?

• Simply plug in $x$ and $y$ in $D(x,y)$ in one step. – Peter Nov 21 '15 at 13:27
• @Peter so since there is no $y$ variable in my $D(x,y)$ do I just plug in $D\left(\frac{4}{3},0\right)$? – hax0r_n_code Nov 21 '15 at 13:28
• This should also work. You have already plugged in $y$. The value for $y$ does not matter anymore. Simply plug in $x$ in the remaining expression. – Peter Nov 21 '15 at 13:29
• @Peter so I plugged this in and got $16$, which means $D\gt0$. I then plugged in $f_{xx}\left(\frac{4}{3}\right)=-8$, which means $f_{xx}\lt 0$ so this means I have a relative maxima? – hax0r_n_code Nov 21 '15 at 13:34
• I do not remember the criterion when we have a maxima. But the plug-in is OK. If you plug in $x$ and $y$ in one step, you should get the same result. Try it ... – Peter Nov 21 '15 at 13:36