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\begin{document}
\vspace*{-0.7true in}

\centerline{\Large \bf The Second Derivative Test for Maxima and 
Minima}

\vskip 20 pt
\noindent{\bf Calculus of One Variable}
\bigskip

In calculus of one variable, if a smooth function $y=f(x)$ has a 
critical point at $x_0$, so $f'(x_0)=0$, one often uses the second 
derivative to test if this critical point is a local maximum or 
minimum.  To understand this, one uses a Taylor polynomial centered 
at $x=x_0$ for $f(x)$, along with the fact that $f'(x_0)=0$: 
 \begin{eqnarray} \label{eq:taylor1}
 f(x) & = & f(x_0) + f'(x_0)(x-x_0) + \tfrac{1}{2} 
           f''(x_0)(x-x_0)^2 + \cdots \hbox{\rm \ higher order terms} \\
  & = &f(x_0) + \tfrac{1}{2}f''(x_0)(x-x_0)^2 + \cdots 
              \hbox{\rm \ higher order terms}.
 \end{eqnarray}
From this one sees that if $f''(x_0)>0$, then $f(x)> f(x_0)$ for all 
$x$ sufficiently near $x_0$ ($x \ne x_0$).  To make this clearer, one can
replace (\ref{eq:taylor1}) by a similar result using The error term 
in Taylor polynomials:
 \begin{equation}\label{eq:taylor2}
 f(x) = f(x_0) + \tfrac{1}{2}f''(z)(x-x_0)^2,
 \end{equation}
for some value of $z$ between $x$ and $x_0$.  If $f''(x)$ is 
continuous and positive at $x_0$, and if $x$ is sufficiently near 
$x_0$, then $f''(z)>0$.  It is now obvious from (\ref{eq:taylor2}) 
that $f$ has a {\it strict local minimum at\/} $x_0$, that is, $f(x) 
> f(x_0)$ for all $x$ sufficiently near $x_0$ ($x\ne x_0$). 

Similarly, if $f''(x_0) < 0$, then $f$ has a strict local minimum at 
$x_0$.  

If $f''(x_0)=0$, then this test fails and one must use higher 
order terms in the Taylor polynomial.  The following graphs
illustrate this.  For each of these the origin is the only critical 
point and the second derivative there is zero.
\bigbreak

\vskip -1.3cm 
\hbox{\hspace*{-0.9cm}\vbox{\epsfig{figure=x3.eps,scale=0.70}} 
\hskip -4.0in \raise 0.0cm \vbox{\epsfig{figure=x4.eps,scale=0.70}}
\hskip -4.0in \raise 0.0cm\vbox{\epsfig{figure=mx4.eps,scale=0.70}}
}
\vskip -1.3cm
$\hskip 1.2cm y=x^3
\hskip 4.2cm  y=x^4
\hskip 4.0cm y=-x^4$


\vskip 20 pt
\noindent{\bf Calculus of Several Variables}
\bigskip

The story for a function of several variables 
$z=f({\bf X})=f(x_1,\ldots,x_n)$, is similar.  Indeed, we will reduce it 
to the one variable case.  Say we have a function $f({\bf X})$ which we 
may imagine to be the temperature at ${\bf X}$.  If 
${\bf X}={\bf P}=(p_1,\ldots,p_n)$ 
is a critical point for $f$ then the first derivative test for 
critical points tells us that $\nabla f({\bf P})=0$.  To see this, let 
${\bf V}=(v_1,v_2,\ldots,v_n)$ be a unit vector at ${\bf P}$ and consider the 
function of {\it one\/} variable 
 $$
 \varphi(t) = f({\bf P}+t{\bf V}),
 $$
which we may thing of as the temperature at the point ${\bf P}+t{\bf V}$ on 
the straight line through ${\bf P}$ in the direction of the vector ${\bf V}$.  
If $f$ has a local minimum at ${\bf P}$, then $\varphi(t)$ has a local
minimum at $t=0$ for every vector ${\bf V}$.  Thus 
$\varphi'(t)|_{t=0} = 0$.  But using the chain rule we have
 \begin{eqnarray}
  \varphi'(t) & = & \frac{d f({\bf P} +t{\bf V})}{dt}
                =   \frac{df(p_1+tv_1,\,\ldots,p_n+tv_n)}{dt} \nonumber \\
              & = & \frac{\partial f({\bf P} +t{\bf V})}{\partial x_1}v_1
           + \cdots + \frac{\partial f({\bf P} +t{\bf V})}{\partial x_n}v_n
\label{eq:chain1} \\
              & = & \nabla f({\bf P}+t{\bf V})\cdot {\bf V}.\nonumber
 \end{eqnarray}
This is of course the usual formula for the directional derivative 
of $f$ in the direction of the vector ${\bf V}$.  In particular, at $t=0$ 
we have
 $$
 0 = \varphi'(0) = \nabla f({\bf P})\cdot {\bf V}.
 $$
Since this holds for all possible vectors ${\bf V}$ we conclude that 
$\nabla f({\bf P})=0$, as claimed.
\medbreak

To obtain the second derivative test we again use the chain rule and 
compute $\varphi''(t)$.  For simplicity, assume $n=2$; the general 
case is essentially identical.  From the first term in 
(\ref{eq:chain1}) we find 
 \begin{eqnarray}                            
  \left(\frac{d}{dt}\right)
  \left(\frac{\partial f({\bf P} +t{\bf V})}{\partial x_1}v_1\right) & = &
  \frac{\partial}{\partial x_1}
      \left(\frac{\partial f({\bf P} +t{\bf V})}{\partial x_1}\right)v_1^2
 + \frac{\partial}{ \partial x_1}
  \left(\frac{\partial f({\bf P} +t{\bf V})}{\partial x_2}\right)v_1v_2, 
\nonumber \\
   & = & f_{11}({\bf P}+t{\bf V})v_1^2 + f_{12}({\bf P}+t{\bf V})v_1v_2,
 \end{eqnarray}
where we have used the standard notation $f_{11}= \partial^2f/\partial x_1^2$, 
etc. There is a similar formula for 
$({d/dt})(\partial f({\bf P} +t{\bf V})/ \partial x_2)$. Thus at $t=0$, after 
we collect terms and use the fact that for a smooth function 
$f_{12}=f_{21}$, we conclude with the formula 
 \begin{equation}
 \varphi''(0)  =  f_{11}({\bf P})v_1^2 + 2f_{12}({\bf P}+t{\bf V})v_1v_2 + 
f_{22}({\bf P})v_2^2.
 \end{equation}
This is a quadratic polynomial in the components of the vector ${\bf V}$ 
and suggests that we write it using the notation developed in the 
previous section for quadratic polynomials in several variables.
Thus we introduce the {\it second derivative matrix\/} (this is 
sometimes called the {\it hessian matrix\/}), 
 $$
  f''({\bf P})= \left(\!\! \begin{array}{cc}
               f_{11}({\bf P}) & f_{12}({\bf P})  \\
               f_{21}({\bf P}) & f_{22}({\bf P})
 \end{array}\!\right).
$$
This gives the cleaner expression
 \begin{equation}\label{eq:deriv2}
 \varphi''(0) = {\bf V}\cdot f''({\bf P}){\bf V}.
 \end{equation}
\smallbreak

Armed with this we can write the Taylor polynomial for $\varphi$ 
about $t=0$
 \begin{eqnarray}
  f({\bf P}+t{\bf V}) & = & \varphi(t) = \varphi(0) + \varphi'(0)t 
  + \tfrac{1}{2}\varphi''(0)t^2 + \hbox{\rm \ higher order terms} \\
  & = & f({\bf P}) + \tfrac{1}{2}[{\bf V}\cdot f''({\bf P}){\bf V}]t^2 +
            \hbox{\rm \ higher order terms}.
  \end{eqnarray}

It should be clear from this that if ${\bf V}\cdot f''({\bf P}){\bf 
V}>0$ for all directions ${\bf V} (\ne 0)$ then for all ${\bf X}={\bf 
P}+t{\bf V}$ near ${\bf P}$, that is, for $t$ sufficiently small, we have  
$f({\bf X}) > f({\bf P})$.  Thus  $f$ has a strict local minimum at 
${\bf P}$.  Similarly if ${\bf V}\cdot f''({\bf P}){\bf V} < 0$ for 
all directions ${\bf V} (\ne 0)$ then  $f$ has a strict local maximum at 
${\bf P}$, while if ${\bf V}\cdot f''({\bf P}){\bf V}$ assumes both 
positive and negative values for various vectors ${\bf V}$, then $f$ has a 
saddle point at ${\bf P}$. 

We can summarize this using the language of the preceding section. 

\bigskip

\noindent{\sc Second Derivative Test} 
{\em Say a smooth function $f$ has a critical point at ${\bf P}$}
$$
 \begin{array}{ll}
\hbox{\rm if\quad } f''({\bf P}) \hbox{\rm\quad is positive definite}\qquad & 
 \hbox{\rm then\ } f \hbox{\rm \quad has a strict local minimum at } {\bf P} \\
\hbox{\rm if\quad } f''({\bf P}) \hbox{\rm \quad is negative definite}\qquad & 
 \hbox{\rm then\ }  f \hbox{\rm \quad has a strict local maximum at } {\bf P} \\
\hbox{\rm if\quad } f''({\bf P}) \hbox{\rm \quad is indefinite}\qquad & 
 \hbox{\rm then\ }  f \hbox{\rm \quad has saddle point at } {\bf P}      \\
\hbox{\rm if\quad } f''({\bf P}) \hbox{\rm \quad is not invertible} \qquad & 
 \hbox{\rm then\ }  f \hbox{\rm \quad no conclusion unless \ } f''({\bf P})
 \hbox{\rm \ is indefinite} 
 \end{array}
$$
The last item above is because if $f''({\bf P})$ is not invertible, 
then the matrix $f''({\bf P})$ can be neither positive definite nor 
negative definite (see {\sc Observation 4} in the preceding 
section).  This is analogous to the situation for a function of one 
variable when the second derivative is zero.  In Examples 3 and 4 
below we will see examples where in these circumstances the behavior 
is uncertain. \bigbreak 

\noindent{\sc Example 1}. Find and classify all the critical points 
of $f(x,y) = x^4 + y^4 -4xy +1$.
\smallskip

\noindent{\sc Solution}. We first find the critical points by solving
 $$
 0 = f_x = 4x^3 -4y \qquad \hbox{\rm and} \qquad 0=f_y= 4y^3-4x.
 $$
We use the first equation $y=x^3$ in the second to find $(x^3)^3-x=0$, that is,
 $$
 0=x^9-x = x(x^8 -1)=x(x^4-1)(x^4+1)=x(x^2-1)(x^2+1)(x^4+1).
 $$
Thus there are three real roots, $x=0, 1, -1$.  Using $y=x^3$ again 
we find there are three critical points: ${\bf P}_1=(0,\,0)$, 
${\bf P}_2=(1,\,1)$, and ${\bf P}_3=(-1,\,-1)$. 

We next calculate the second derivative matrix.
 $$
  f''({\bf P})= \left(\!\! \begin{array}{cc}
               \phantom{-4}12x^2 & \phantom{12x}-4\phantom{1^2}  \\
               \phantom{12x}-4\phantom{1^2} & \phantom{-4}12y^2
 \end{array}\!\right),
$$
so that at ${\bf {\bf P}}_1,\ {\bf P}_2,\ {\bf P}_3$
 $$
  f''({\bf P}_1)= \left(\!\! \begin{array}{cc}
                \phantom{-}0 & -4  \\
                          -4 & \phantom{-}0
 \end{array}\!\right), \qquad
  f''({\bf P}_2)= \left(\!\! \begin{array}{cc}
                \phantom{-}12 & \phantom{}-4  \\
                \phantom{}-4 & \phantom{-}12
 \end{array}\!\right), \quad \hbox{\rm and}\quad
  f''({\bf P}_3)= \left(\!\! \begin{array}{cc}
                \phantom{-}12 & \phantom{}-4  \\
                \phantom{}-4 & \phantom{-}12
 \end{array}\!\right).
$$
\vskip -1.8cm

\hbox{\begin{minipage}{4.1in}
Using the determinant test of the previous section, we see that 
$f''({\bf P}_1)$ is indefinite, while both $f''({\bf P}_2)$ and 
$f''({\bf P}_3)$ are positive definite.  Consequently, $f$ has a 
saddle point at ${\bf P}_1$ and strict local minima at ${\bf P}_2$ 
and ${\bf P}_3$.  The graph illustrates this clearly. 
\end{minipage}
\hskip -2.0cm \lower 1.6in\vbox{\epsfig{figure=mms.eps,scale=1.0}}
}

\vskip -2.0cm

\noindent{\sc Example 2}. Find and classify all the critical points of
 $$
 h(x,y)= \frac{3x^4 + 4x^3 -12x^2 +6}{12(1+y^2)}.
 $$

\noindent{\sc Solution}.  To find the critical points we must solve
 $$
 0 = h_x = \frac{x^3 + x^2 -2x}{1+y^2} \quad and \quad
 0 = h_y = \frac{-y(3x^4 +4x^3 - 12x^2 +6)}{6(1+y^2)^2}.
 $$
Since $x^3 + x^2 -2x=x(x^2+x-2)=x(x-1)(x+2)$, the first equation is 
satisfied only if $x=0,\,1$, or $-2$.  Since none of these 
satisfy $3x^4 +4x^3 - 12x^2 +6=0$, the only way the second equation, 
$h_y=0$, is satisfied is if $y=0$.  We thus obtain three critical 
points ${\bf P}_1=(0,0), \ {\bf P}_2=(1,0)$, and ${\bf P}_3=(-2,0)$. 

To classify these critical points we need the second derivative 
matrix.  It is
 $$
 f''({\bf P})= \left(\!\! \begin{array}{ccc}
    \frac{3x^2+2x-2}{ 1+y^2}    &\frac{\-2y(x^3+x^2-2x)}{(1+y^2)^2} \\
    \frac{\-2y(x^3+x^2-2x)}{(1+y^2)^2}  & 
    \frac{-(1-3y^2)(3x^4 + 4x^3 -12x^2 +6)}{ 6(1+y^2)^3}.
 \end{array}\!\right).
 $$
Thus
 $$
 f''({\bf P}_1)= \left(\!\! \begin{array}{cc}
                          -2 & \phantom{-}0  \\
                \phantom{-}0 & -1
 \end{array}\!\right), \qquad
  f''({\bf P}_2)= \left(\!\! \begin{array}{cc}
                          3 & \phantom{-}0  \\
                          0 & -\tfrac{1}{6}
 \end{array}\!\right), \quad \hbox{\rm and}\quad
  f''({\bf P}_3)= \left(\!\! \begin{array}{cc}
                6 & \phantom{}0 \\
                0 & \tfrac{13}{3}
 \end{array}\!\right). 
 $$
\vskip -1.2in
\hbox{\begin{minipage}{4.0in}
Using {\sc Observations} 1 and 3 of the preceding section (or the 
determinant test) we see that $f''({\bf P}_1)$ is negative definite, 
$f''({\bf P}_2)$ is indefinite, while $f''({\bf P}_3)$ is positive 
definite.  Thus ${\bf P}_1$ is a strict local maximum, ${\bf P}_2$ a 
saddle point, and ${\bf P}_3$ a strict local minimum.  The graph 
clarifies this. 
\end{minipage}
\hskip -2.6cm \lower 1.8in\vbox{\epsfig{figure=Mms.eps,scale=1.2}}
}

\vskip-1.1in

\noindent{\sc Example 3}. The functions
 $$
 f(x,y) = x^2 - y^3 \qquad g(x,y) = x^2 +y^4 \qquad h(x,y)= x^2 -y^4
 $$
each have only one critical point, located at the origin.  For each of 
them the second derivative matrix there is
 $$
  \left(\!\! \begin{array}{cc}
                          2 & 0  \\
                          0 & 0
 \end{array}\!\right).
 $$
This matrix is not invertible and is not indefinite, so the second 
derivative test gives no information.  To analyze the nature of 
these critical points one must either investigate the higher order 
terms in the Taylor polynomial at the origin, or else look at a 
graph.  Either way it is clear that the origin is a more exotic 
saddle point for the first example, a strict local minimum for the 
second, and a saddle for the third. \smallskip 

\vskip -1.1cm 
\hbox{\hspace*{-1.2cm}\vbox{\epsfig{figure=flatcubic.eps,scale=0.80}} 
\hskip -4.0in \raise 0.0cm \vbox{\epsfig{figure=flatmin.eps,scale=0.80}}
\hskip -4.0in \raise 0.0cm\vbox{\epsfig{figure=flatsaddle.eps,scale=0.80}}
}
\vskip -1.3cm
$\hskip 0.9cm  z = x^2 - y^3
\hskip 3.2cm  z = x^2 + y^4
\hskip 3.1cm  z = x^2 + y^4$

\noindent{\sc Example 4}. The examples $p(x,y)=x^3 -3xy^2$ (a {\it
monkey saddle\/}, since there is also a place for its tail),
$q(x,y)=x^4 + y^4$, and $r(x,y)=-x^4 - y^4$, all of which have the
origin as their only critical point and have second derivative matrix
$0$, further show that if at a critical point the second derivative
matrix is not invertible, the nature of the critical point is
determined only by a more thorough study.

\vskip -1.1cm 
\hbox{\hspace*{-1.2cm}\vbox{\epsfig{figure=monkey.eps,scale=0.80}} 
\hskip -3.9in \raise -0.3cm \vbox{\epsfig{figure=FlatsaddleU.eps,scale=0.80}}
\hskip -4.0in \raise 0.1cm \vbox{\epsfig{figure=FlatsaddleD.eps,scale=0.80}}
}
\vskip -1.3cm
$\hskip 0.7cm  z = x^3 - 3xy^2
\hskip 3.3cm  z = x^4 + y^4
\hskip 3.2cm  z = -x^4 -  y^4$

\bigbreak

\noindent{\sc Example 5}. The function $F(x,y)=x^2 +2xy+y^2$ has critical 
point where both
 $$
 0 = F_x = 2x +2y \quad \hbox{\rm and} \quad 0=F_y = 2x + 2y.
 $$
that is, along the whole line $y=-x$.  The second derivative matrix there is
 $$
  \left(\!\! \begin{array}{cc}
                          1 & 1  \\
                          1 & 1
 \end{array}\!\right),
 $$
\vskip -2.7cm
\hbox{\begin{minipage}{4.4in}
which is not invertible, so the second derivative test fails.  However, 
either from the observation that we can write $F(x,y)=(x+y)^2$ or else from 
the graphs, we conclude that these critical points are all local minima -- 
but not strict local minima.
\end{minipage}
\hskip -1.3cm \lower 1.0in\vbox{\epsfig{figure=trough.eps,scale=0.80}}
}


\centerline{\bf Problems}
\parindent=0pt
\bigskip

1. Find and classify the critical points of the function in 
Problem 2 of the preceding section. 

\bigbreak


2. Find and classify the critical points of the following
functions -- even if the second derivative test is not applicable.
\smallskip

\hbox{\begin{minipage}{3.0in}
\begin{parts}
\item $f(x_1,x_2) = x^2_1 + 4x_1 + 2x^2_2 + 10$
\item $\varphi (x,y) = 3 - 2x + 2y + x^2y^2$
\item $f(x,y) = xy - x + y - 2$
\item $u(x_1,x_2) = x^2_1 - x^2_2$
\item $g(x,y) = y^3 - 2x^2 - 2y^2 + y$
\item $v(x,y) = (x^2 + y^2)^2 - 8y^2$
\item $\psi (x_1,x_2) = \dfrac{\cos x_2}{1+x^2_1}$
\item $h(x,y) = \dfrac{\cosh (2x)}{1+2y^2}$
\end{parts}
\end{minipage}
\begin{minipage}{3.0in}
\begin{parts}\addtocounter{bean}{8}
\item $w(x,y) = x^2 - 2xy + \frac{1}{3}y^3 - 3y$
\item $\psi (y,z) = \dfrac{z^3 - 3z}{1+y^2}$
\item $k(x_1,x_2) = (x^2_1 - 1)e^{x_{2}}$
\item $g(x,y,z) = \dfrac{xye^y}{1+x^2} + z^2 + 2z$
\item $h(x,y,z) = \dfrac{3x^4 - 6x^2 + 1}{1+z^2} - y^2$
\item $\varphi (x,y,z) = x^3 - 3x + y^2 + z^2$
\item $r(x,y,z) = (1 + 2x + 3y - z)^2\;.$
\end{parts}
\end{minipage}
}

\bigbreak

\hbox{\begin{minipage}{4.2in}
3. Find and classify the critical points of the function
 $$
 z=z(\phi,\,\theta) = (2+\cos\phi)\sin\theta,\qquad
        0 \le \theta,\phi <2\pi.
 $$
This is the $z$ coordinate of a torus standing on its edge; its other
coordinates are $y=(2+\cos\phi)\cos\theta$ and $x= \sin\phi$ (see
fig).
\end{minipage}
\hskip -1.2cm \lower 1.2in\vbox{\epsfig{figure=torus.eps,scale=0.8}}
}

\vskip -1.3cm

4. Find and classify the critical points of the following functions.  
For the first two, also draw graphs showing the critical points.
\begin{parts}
\item $h(x,y) = (x^2 +2y^2)e^{1-(x^2+y^2)}$.
\item $k(x,y) = (x^2 -2y^2)e^{1-(x^2+y^2)}$.
\item $q(x,y,z) = (x^2 + 2y^2 + 3z^2)e^{1-(x^2+y^2+z^2)}$
\end{parts}

\bigbreak


5. Find and classify the critical points of both of the functions
\begin{parts}
\item $q(x,y,r,s)= 5x^2  + 2xy - 6xs + 5y^2 - 6ys - 9s^2 - 18xr - 18yr + 
3r^2 + 36rs $,
\item $h(x,y,r,s)= 5x^2  + 2xy - 6xs + 5y^2 - 6ys + 9s^2 - 18xr - 18yr + 
3r^2 + 36rs$.
 \end{parts}

 \bigbreak


6. This exercises concerns three games.
\begin{parts}
\item Susan picks any real number $x$ that she wants.  Then, Gwen, 
after hearing Susan's choice, picks a real number $y$ of her choice 
and pays Susan $xy-3x+2y$ dollars.  What number should Susan pick? 
(A negative value of $xy-3x+2y$ indicates a payment from Susan to 
Gwen.)  A graph of the ``payoff function'' $xy-3x+2y$ may help 
understanding.
\item Discuss the modified game in which Gwen  chooses first and, 
then, Susan makes her choice after hearing Gwen's choice. 
\item Discuss the modification in which each player writes her 
choice secretly on a piece of paper, before hearing the choice of 
the other player.
\item Use Maple to simulate these games, with the computer playing 
the role of Susan and you playing the role of Gwen.  Have the 
computer make random choices and you make whatever choice you find 
appropriate.  Tabulate the ``payoff'' to understand which strategies 
are optimal.  
 \end{parts} 

 \bigbreak


7. Let $A$ be a symmetric matrix and let $f$ be the quadratic 
polynomial $f({\bf X})={\bf X}\cdot A{\bf X}$.
\begin{parts} 
\item Show that ${\rm grad}\,f({\bf X})= A{\bf X}$ and 
that $f''({\bf X}) = 2A$.  Note the similarity between this and the 
special case of a function of one variable case $f(x)=ax^2$ where 
$f''(x)=2a$. 
\item  If $A$ is invertible, show that the only critical 
point of $f$ is at the origin, and that this is a local minimum if 
and only if $A$ is positive definite. 
\item  Extend parts a) and b) to the more general case where the 
quadratic polynomial has lower order terms, 
$f({\bf X})={\bf X}\cdot A{\bf X} + 2{\bf b}\cdot{\bf X} + c$, where 
${\bf b}$ is a vector and $c$ a scalar.  As a check, compare your 
result with that obtained in the special case of one variable, 
$f(x)=ax^2 +2bx +c$.
\end{parts}

\bigbreak


8. {\sc Least Squares.}  Let $A$ be a matrix, not necessarily square, 
$b$ a given vector, and let $f$ be the quadratic polynomial 
$f({\bf X})=\|A{\bf X} -{\bf b}\|^2$.  The problem is to minimize 
$f$.  If $A$ is invertible, just let ${\bf X}$ be the solution of 
$A{\bf X}={\bf b}$.  The difficulty is if $A$ is not invertible, say 
because there are more equations than unknowns.  Then $f({\bf X})$  
measures the discrepancy between $A{\bf X}$ and ${\bf b}$.  The 
optimal ${\bf X}$ is the value that minimizes this discrepancy. This 
problem arises, for instance, when one tried to fit experimental 
data to a straight line that one does not expect to pass through all 
of the points. 
 \begin{parts} 
\item  Show that $\hbox{\rm grad}\, f({\bf X})= 
2A^T(A{\bf X}-{\bf b})$. 
\item  Compute the second derivative matrix, $f''$, and 
show that if the only solution of the homogeneous equation $A{\bf 
Z}=0$ is $Z=0$, then $f$ has exactly one critical point and it is a 
local minimum. 
\item  Use this to find the ``optimal'' solution of the 
following set of four equations in two unknowns:
 \begin{eqnarray*}
        2x + 3y & =  \phantom{-}1 \\
        3x + 3y & =  \phantom{-}2 \\
        2x + 2y & =  \phantom{-}0 \\
         x      & =            -1
  \end{eqnarray*}
\end{parts}

\bigbreak


9. If $f(x,y)$ has a local minimum at the point $(a,\,b)$, show that 
$f''(a,b)$ must be positive definite or semi-definite. [Suggestion: 
First state and prove the version for a function of one variable].

\bigbreak


10. If $u(x,y)$ has the property that $2u_{xx} + 3u_{yy}= -1$, show that 
$u$ cannot have a local minimum {\it anywhere\/}.

\bigbreak


11. If the function $f(x,y)$ has the property that $f''(x,y)$ is 
positive definite or semi-definite at every point $(x,\,y)$, show 
that the surface $z=f(x,y)$ lies above its tangent plane at every 
point, that is, the surface is {\it convex\/}.  [Suggestion: First 
try the analogous assertion for a function of one variable.] 

\bigbreak


12. Let $k(x,y)=(y-x^2)(y-2x^2)$.
\begin{parts} 
\item Show that the origin is the only critical point.
\item  Show that $k(x,y) > 0$ for all points $(x,y)$ 
except those in the ``horn'' region $x^2\le y \le 2x^2$, where 
$k(x,y) \le 0$.  Thus the origin is a saddle point. 
\item  On the other hand, note that moving from the origin along 
straight lines the function is positive for at least a short 
distance.  Since $k(0,0)=0$, as one approaches the origin along 
straight lines, the origin appears to have a local maximum.  The 
point is that because the second derivative matrix $k''(0,0)$ is not 
invertible, the behavior of $k$ near the origin is too complicated 
to be resolved by just considering the behavior of $k$ along 
straight through the origin. [However, had $k''(0,0)$ been 
invertible, then straight lines would have been adequate].
 \end{parts}

 \bigbreak


13. Let $A$ be an invertible symmetric matrix and let 
 $ f({\bf X})= ({\bf X}\cdot A{\bf X})\,e^{-\|X\|^2}$.
\begin{parts} 
\item Show that the critical points of $f$ are exactly 
the origin and all the unit vectors that are eigenvectors of $A$. 
\item  In addition assume that $A$ is a positive definite 
$n \times n$ matrix with {\it distinct\/} eigenvalues $0<\lambda_1 < 
\lambda_2 < \ldots \lambda_n$ and corresponding orthonormal 
eigenvectors ${\bf V}_1,\ldots,{\bf V}_n$. Show that $f$ has a 
strict local maximum at the two points $\pm{\bf V}_n$, a strict 
local minimum at the origin, while the remaining $2(n-1)$ critical 
points are saddle points. [Suggestion: first understand the case 
when $A$ is a diagonal matrix].
 \end{parts} 

 \end{document}
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