# Maxima, minima, and saddle points

I've got the following function $$f(x,y)=x+y-xy$$ where $x,y\in \mathbb R$. I need to determine whether there is minima, maxima or saddle point. Easily enough, after doing the partial derivatives

$f_x=1-y$

$f_y=1-x$

then $c=(1,1)$ is critical point. Let $H$ denote the "Hessian" matrix of second partial derivatives, then Hessian matrix of $f$ at c as the following $$H(f(c))=\begin{bmatrix} 0 & -1 \\ -1 & 0 \end{bmatrix}$$

but the Hessian matrix of this function at the origin is the null matrix, which is not indefinite.

How can I prove mathematically that $c=(1,1)$ is a saddle point.

I know the third condition is satisfied. But, how can I prove mathematically that $c=(1,1)$ is a saddle point.

If $detH(f(c)) < 0$, then $f(x,y)$ takes on both positive and negative values in a neighborhood $c$?

Please I need some references on this type and what the appropriate test if Hessian matrix is not indefinite.

• Sorry, I don't understand what the trouble is in applying the second derivative test here. You have calculated the Hessian matrix, and if you compute the determinant you should get that it is negative, so then you are done, right? – Joey Zou Sep 3 '17 at 10:47
• @JoeyZou This is a simple example, we have some examples that the test fails to explain, I need mathematical proof, this test is certainly built on a mathematical proof . For example, $f(x,y)=x^3y^3$ the test fails to determine the type of critical point of $f$. – Emad kareem Sep 3 '17 at 10:55
• I see. In that case to show that a critical point is a saddle point, you need to show that it is neither a maximum nor a minimum. One way to show that is to find two curves through your critical point (for example, two lines through your critical point) where your function doesn't attain a max at the critical point along one of the curves, while the opposite is true on the other curve. – Joey Zou Sep 3 '17 at 11:05
• @JoeyZou That's exactly what I want. Can you make an answer to a function in the question, as stated in the commentary with the explanation. – Emad kareem Sep 3 '17 at 11:11

Hint. Note that $$f(x,y)=1-(x-1)(y-1).$$ which implies that $f(x,y)-1$ changes its sign in a neighbourhood of the critical point $(1,1)$ and therefore it is not a local extremum. What may we conclude?

P.S. Recall the definition of saddle point.

• This proof is not clear to me, can you explain more. – Emad kareem Sep 3 '17 at 10:11
• We have that $f(x,y)<1$ in the quadrant $\{(x,y): x>1 , y>1\}$, and $f(x,y)>1$ in the quadrant $\{(x,y): x>1 , y<1\}$. Therefore $(1,1)$ is a critical point but it is not a local maximum/minimum. – Robert Z Sep 3 '17 at 10:14
• Can I supposed that the domain of $f$ is the whole $\mathbb R^2$. then I study either $$x+y-xy>0$$ or $$x+y-xy<0$$ With an $(x,y)$ plot and focusing on what happens around $(1,1)$ i.e. there exists no such neighborhood $U$, so $(1,1)$ cannot be either a max or a min). – Emad kareem Sep 3 '17 at 10:23
• @Emad kareem You should study $x+y-xy>f(1,1)=1$ and $x+y-xy<f(1,1)=1$. – Robert Z Sep 3 '17 at 10:25
• My problem I do not have the information and experience about this type of proof. Can you give me some references with thanks and appreciation to you. – Emad kareem Sep 3 '17 at 10:29

Let $\gamma_1,\gamma_2:\mathbb{R}\rightarrow\mathbb{R}^2$ be defined by $\gamma_1(t) = (1+t,1+t)$ and $\gamma_2(t) = (1+t,1-t)$, and let $g_1 = f\circ\gamma_1$, $g_2 = f\circ\gamma_2$. Notice that $\gamma_1(0) = \gamma_2(0) = (1,-1)$, and \begin{align} g_1(t) &= (1+t)+(1+t)-(1+t)(1+t) = 1-t^2 \\ g_2(t) &= (1+t)+(1-t)-(1+t)(1-t) = 1+t^2. \end{align} We thus see that $g_1$ attains a strict maximum at $t=0$, while $g_2$ attains a strict minimum at $t=0$. This means that $f$, restricted to the curve $\gamma_1$, attains a strict maximum at $(1,-1)$, while $f$ restricted to the curve $\gamma_2$ attains a strict minimum instead. Thus, $(1,-1)$ is a saddle point for $f$, as it is a critical point which cannot be a local maximum nor a local minimum.

How did I come up with the $\gamma_1,\gamma_2$? Basically, these are curves passing through the critical point $(1,1)$ at $t=0$ whose derivatives at $t=0$, namely $(1,1)$ and $(1,-1)$, are the eigenvectors of the Hessian $H(1,1) = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}$, of eigenvalues $-1$ and $1$, respectively. Why should these curves work? By Taylor's formula, we have \begin{align} f(x_0+x)-f(x_0) &\approx (\nabla f(x_0))^Tx + \frac{1}{2}x^TH(x_0)x \\ &=\frac{1}{2}x^TH(x_0)x\quad\text{if }\nabla f(x_0) = 0\text{, i.e. }x_0\text{ is a critical point.} \end{align} Thus if $x = t\begin{pmatrix} 1 \\ 1 \end{pmatrix}$, then $x$ is an eigenvector of $H$ of eigenvalues $-1$, and hence $$\frac{1}{2}x^THx = \frac{1}{2}x^T(-x) = -\frac{1}{2}\left(t\begin{pmatrix} 1 \\ 1 \end{pmatrix}\right)^T\left(t\begin{pmatrix} 1 \\ 1 \end{pmatrix}\right) = -t^2$$ while similarly $x = t\begin{pmatrix} 1 \\ -1\end{pmatrix}\implies \frac{1}{2}x^THx = t^2$. Since $\gamma_1(t) = (1,1)+t(1,1)$ (and similarly for $\gamma_2$), by our arguments we have $$f(\gamma_1(t))-f(1,1) \approx -t^2 <0$$ and $$f(\gamma_2(t))-f(1,1) \approx t^2 > 0.$$ To make the arguments fully rigorous, one has to replace the "$\approx$" in Taylor's formula by the actual error, which produces a term that vanishes faster than quadratically and hence will not affect our estimates. In this case the formula is actually exact, since we are just dealing with a second-order polynomial.