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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

$$0=2\left(\frac{3}{\sqrt{y}}i\right)^3y+27\left(\frac{3}{\sqrt{y}}i\right)+27=\frac{27}{\sqrt{y}}i+27=0,$$

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.

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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

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ 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

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