I have that $f'_x=3x^2y^2+27y$ and $f'_y=2x^3y+27x+27$. Setting $f'_x=0$ I get that

$$x^2=\frac{-27y}{3y^2}=-\frac{9}{y}\Leftrightarrow x=\pm\frac{3}{\sqrt{y}}i$$

Setting $x=\frac{3}{\sqrt{y}}i$ in the equation $f'_y=0$ gives


Which gives me the equivalent equation

$$i+\sqrt{y}=0\Leftrightarrow \sqrt{y}=-i \Leftrightarrow y=(-i)^2=-1.$$

This is a false root however, so for $x=\frac{3}{\sqrt{y}}i$ no roots for $y$ exists. Using $x=-\frac{3}{\sqrt{y}}i$ I get instead $i-\sqrt{y}=0$ and this equation as the root $y=-1$. This means that $x=-3.$ So, one stationary point is $(x,y)=(-3,-1).$ However the book says there is another stationary point, namely $(-1,0).$ However I've found all values for $y$ and the only working value that makes sense is $y=-1$, how can I find $y=0$? It doesn't make sense for $y$ to be $0$ since that would mean division by zero when I want to get my x-value.


you must solve the system $$\frac{\partial f(x,y)}{\partial x}=3x^2y^2+27y=0$$ and $$\frac{\partial f(x,y)}{\partial y}=2x^3y+27x+27=0$$ simultaneously. solving this we get $$x=-3,y=-1$$ or $$x=-1,y=0$$ from the first equation we get $$y=0$$ or $$x^2y=-9$$ so $$y=-\frac{9}{x^2}$$ and plug this in the second one

  • $\begingroup$ Which is what I did. I solved for $x$ in the first and plugged it into the second one. $\endgroup$ – Parseval Feb 2 '18 at 15:13
  • $\begingroup$ and what are your results? $\endgroup$ – Dr. Sonnhard Graubner Feb 2 '18 at 15:17
  • $\begingroup$ The above. I only get $(x,y)=(-3,-1)$. How did you solve this simultaneous equation? $\endgroup$ – Parseval Feb 2 '18 at 15:30

Here a way to solve the system $\;\begin{cases}3x^2y^2+27y=0,\\[1ex]2x^3y+27x+27=0.\end{cases}$

The first equation is equivalent to $\;y(x^2y+9)=0$. So:

  • either $y=0$, and the second equation yields $\;x=-1$;
  • or $x^2y=-9$. Plugging this relation into the second equation, multiplied by $xy$ (note that $x$ has to be $\ne 0$): $$2x^4y^2+27x^2y+27xy=0 \Rightarrow \underbrace{2\cdot 81}_{6\mkern1mu\cdot 27}-27\cdot 9+27xy=0\Leftrightarrow 27xy= 3\cdot 27\Leftrightarrow xy=3.$$ Going back to the first equation, we get $\;x^2y=3x=-9$, whence $\;x=-3 $, then $\;y=-1$.

Thus, there are two candidates: $(-1,0)$ and $(-3,-1)$.

Now calculate the hessian: $$H_f=\begin{vmatrix}f''_{x^2}&f''_{xy}\\f''_{yx}&f''_{yx^2}\end{vmatrix}= \begin{vmatrix}6xy^2& 6x^2y+27\\ 6x^2y+27 & 2x^3\end{vmatrix}=\begin{cases} {\scriptstyle\begin{vmatrix}0&27\\27&-2\end{vmatrix}}=-729,\\[1ex] {\scriptstyle\begin{vmatrix}-18&-27\\-27&-2\cdot 27\end{vmatrix}}=9\cdot27\,{\scriptstyle\begin{vmatrix}2&1\\3&2\end{vmatrix}}=243. \end{cases}$$

As the hessian is negative in the first case, we have a saddle point.

In the second case, it is positive, so we have a local extremum. Furthermore, the trace of the hessian is negative, which implies this extremum is a local maximum.

  • $\begingroup$ Should one check that that this is a minimum for all directions, and not only for $x$ and $y$ axis? Remember minimum/maximum definition $\endgroup$ – Carlos Campos Feb 2 '18 at 16:50
  • $\begingroup$ No, it's not necessary: it's a minimum because of a theorem which results from Taylor's formula for several variables at order two (the associated quadratic form is definite negative). $\endgroup$ – Bernard Feb 2 '18 at 16:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.