# What are the local extreme values of $f(x,y)=x^3+x^2y-y^2-4y$?

$f'_x=3x^2+2xy$, $f'_y=x^2-2y-4$

so I have to solve the equation system of $3x^2+2xy=0,x^2-2y-4=0$ as solutions I get that $((x=0),(y=0)),((x=-4),(y=6)),(x=1),((y=-\frac{3}{2}))$

After that:

$f''_x=6x+2y$, $f''_y=-2$, $f''_{xy}=2x=f''_{yx}$

so I get the determinant $\begin{vmatrix} 6x+2y & 2x \\ 2x & -2 \end{vmatrix}$

for $(x=4),(y=6)$

$\begin{vmatrix} 6 \cdot -4+2 \cdot6 & 2 \cdot -4 \\ 2 \cdot -4 & -2 \end{vmatrix}$=$-40$ there is at this point no local extreme value

for $(x=1),(y=-\frac{3}{2})$

$\begin{vmatrix} 6 \cdot 1+2 \cdot-\frac{3}{2} & 2 \\ 2 & -2 \end{vmatrix}$=$-10$ there is at this point no local extreme value

but what about $(x=0)(y=0)$? the value of the determinant will be $0$but it says nothing about the extreme value at this point.

Is my solution correct til this point?

• The point (0,0) is not a critical point. This is where your mistake is.
– D.B.
May 3, 2018 at 0:02
• omg really, and is my solution correct with the exception of this $(0,0)$ part? May 3, 2018 at 0:06
• I think so. On wolfram, I see only one extremum at the point (0,-2).
– D.B.
May 3, 2018 at 0:22
• so will be also (0,-2) a solution of the equation system? May 3, 2018 at 0:31

The stationary points are determined as the solutions for

$$3x^2+2xy = 0\\ x^2-2y-4 = 0$$

giving the set

$\{(-4,6),(0,-2),(1,-3/2)\}$

Now taking the Hessian

$$H = \left(\begin{array}{cc}3x+y & x\\ x & -1\end{array}\right)$$

To qualify the stationary points we should evaluate $H$ in such set, verifying it's eigenvalues. For both eigenvalues negative we have a local maximum. For both eigenvalues positive we have a local minimum and for eigenvalues with opposite sign we have a saddle point. In this case the point $(0,2)$ gives negative eigenvalues for $H$ so here we have a local maximum.

Attached a level contour plot showing the local maximum (red) and the two saddle points (blue)