0
$\begingroup$

Determine all local and global extrema of the function \begin{equation*}f:\mathbb{R}^2\rightarrow \mathbb{R} , \ \ f(x,y)=x^3+3xy^2-3x+1\end{equation*} Determine also the type of extrema : minimum, maximum, local, global.

$$$$

I have done the following :

First we calculate the gradient : \begin{equation*}\nabla f=\begin{pmatrix}f_x \\ f_y\end{pmatrix}=\begin{pmatrix}3x^2+3y^2-3 \\ 6xy\end{pmatrix}\end{equation*} Then we set the gradient equal to $0$ and we solve the syste that we get: \begin{equation*}\begin{pmatrix}3x^2+3y^2-3 \\ 6xy\end{pmatrix}=\begin{pmatrix}0\\ 0\end{pmatrix}\Rightarrow \begin{cases}3x^2+3y^2-3=0 \\ 6xy=0\end{cases}\end{equation*} From the second equation we get $x=0$ or $y=0$. If $x=0$ we get from the first equation $3y^2-3=0\Rightarrow y^2=1 \Rightarrow y\pm 1$. If $y=0$ we get from the first equation $3x^2-3=0\Rightarrow x^2=1 \Rightarrow x\pm 1$. So we get the extrema $(0,-1)$, $(0,1)$, $(-1,0)$, $(1,0)$.

Now we determine the Hessian matrix : \begin{equation*}H_f(x,y)=\begin{pmatrix}f_{xx} & f_{xy} \\ f_{yx} & f_{yy}\end{pmatrix}=\begin{pmatrix}6x & 6y \\ 6y & 6x\end{pmatrix}\end{equation*}

  • \begin{equation*}H_f(0,-1)=\begin{pmatrix}0 & -6 \\ -6 & 0\end{pmatrix}\end{equation*} The eigenvalues are $\lambda_1=-6$ and $\lambda_2=6$. The matrix is indefinite. So at $(0,1)$ the function has a saddle point.
  • \begin{equation*}H_f(0,1)=\begin{pmatrix}0 & 6 \\ 6 & 0\end{pmatrix}\end{equation*} The eigenvalues are $\lambda_1=-6$ and $\lambda_2=6$. The matrix is indefinite. So at $(0,1)$ the function has a saddle point.
  • \begin{equation*}H_f(-1,0)=\begin{pmatrix}-6 & 0 \\ 0 & -6\end{pmatrix}\end{equation*} The eigenvalues are $\lambda_1=-6$ and $\lambda_2=-6$. The matrix is negative definite. So at $(-1,0)$ the function has a local maximum.
  • \begin{equation*}H_f(1,0)=\begin{pmatrix}6 & 0 \\ 0 & 6\end{pmatrix}\end{equation*} The eigenvalues are $\lambda_1=6$ and $\lambda_2=6$. The matrix is positive definite. So at $(1,0)$ the function has a local minimum.

Do we have to check also some other points or can we say that at $(-1,0)$ the function has also a global maximum and at $(1,0)$ a global minimum?

$\endgroup$
2
  • $\begingroup$ You can rewrite the function as, $f(x,y) = x (x^2 + 3y^2 - 3) + 1$ and that shows that as $x \to +\infty$, $f(x, y) \to +\infty$ and similarly when $x \to - \infty$, $f(x, y) \to -\infty$ $\endgroup$
    – Math Lover
    Sep 3, 2021 at 14:34
  • $\begingroup$ I see! Thank you very much!! :-) @MathLover $\endgroup$
    – Mary Star
    Sep 3, 2021 at 15:19

1 Answer 1

2
$\begingroup$

That function neither has a global maximum nor a global minimum. Note that $f(x,0)=x^3-3x+1$ and that $\lim_{x\to\pm\infty}x^3-3x+1=\pm\infty$.

$\endgroup$
5
  • $\begingroup$ Ahh ok! So only the local minimum at $(1,0)$ and the local maximum at $(-1,0)$, right? $\endgroup$
    – Mary Star
    Sep 3, 2021 at 14:26
  • 1
    $\begingroup$ Yes, those are just local extreme points. $\endgroup$ Sep 3, 2021 at 14:28
  • $\begingroup$ Ok! In general to check if a local extremum is also a global one, do we always have to set one variable equal to zero and check the behaviour of the function when the other variable goes to infinity? $\endgroup$
    – Mary Star
    Sep 3, 2021 at 14:30
  • 1
    $\begingroup$ I am not aware of a general method that works every single time. But when this method works, it's usually simple to apply. $\endgroup$ Sep 3, 2021 at 14:35
  • $\begingroup$ Ok! Thank you very much!! :-) $\endgroup$
    – Mary Star
    Sep 3, 2021 at 15:19

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.