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

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?

  • 1
    $\begingroup$ The point (0,0) is not a critical point. This is where your mistake is. $\endgroup$
    – D.B.
    May 3, 2018 at 0:02
  • $\begingroup$ omg really, and is my solution correct with the exception of this $(0,0)$ part? $\endgroup$
    – Zauberkerl
    May 3, 2018 at 0:06
  • 1
    $\begingroup$ I think so. On wolfram, I see only one extremum at the point (0,-2). $\endgroup$
    – D.B.
    May 3, 2018 at 0:22
  • $\begingroup$ so will be also (0,-2) a solution of the equation system? $\endgroup$
    – Zauberkerl
    May 3, 2018 at 0:31

1 Answer 1


The stationary points are determined as the solutions for

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

giving the set


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)

enter image description here


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .