\be
\item Determine the quadratic approximation,
 \begin{equation*}
 f(\vecx )= f(\vecx_0 )+ \nabla f(\vecx_0)^t ( \vecx-\vecx_0 )+ \frac{1}{2}(\vecx-\vecx_0)^t H_f(\vecx_0)(\vecx-\vecx_0)
 \end{equation*}
    for the function $\ds f(x_1,x_2)=e^{-\left(x_1^2+x_2^2\right)}$ at
    $\ds \vecx_0=\begin{bmatrix} .5 \\ .5 \end{bmatrix}$. Plot this approximation together with the function $f$.
\item Recall \emph{Pam's Pentathlon Training Program}\index{Pam's Pentathlon Problem}, Exercise 2, from Section \ref{S:NLP_Overview}.
When stated as a maximization problem, the objective is represented by the
function $f:S \to \R$, where
$\ds S=\R_+^3=\left\{(x_1,x_2,x_3) \mid x_1, x_2,x_3 \geq 0 \right\}$ and where
\begin{align*}
f(x_1,x_2,x_3)&= .11193(254-(180-.5x_1-x_2-x_3))^{1.88} \\
& \qquad \quad + 56.0211(5.2+.1x_1)^{1.05}.
\end{align*}
In this formula, $x_1$, $x_2$, and $x_3$ denote the number of hours each week Pam devotes
to weight lifting, distance running, and speed workouts. The value of $f(x_1,x_2,x_3)$
represents the portion of Pam's total pentathlon score due to her performances in the 800 meter
run and the shot put.

\noi
Show that the function $f$ is strictly convex on $S$ by establishing that its Hessian is
positive definite there. (Hint: Express the Hessian in the form
$\ds H_f(\vecx)=\phi(\vecx) A$, where $\phi$ is a positive function defined on $S$ and where
$A$ is a positive definite matrix of numbers. Then each eigenvalue of $\ds H_f(\vecx)$ is a product
of $\phi(\vecx)$ and an eigenvalue of $A$.)

\item Calculate all critical points of each of the following functions on its
stated domain $S$. Determine whether the function is convex, concave, or neither on $S$ and then
classify each critical point as a local or global minimum, local or global maximum, or saddle point.
	\be
	\item $\ds f(x_1,x_2)=x_1^2+x_1x_2-x_1+2x_2^2-4x_2+2, \quad S=\R^2$
    \item $\ds f(x_1,x_2)=x_1^2-10x_1x_2+4x_1+7x_2^2-8x_2+2, \quad S=\R^2$
     \item $\ds f(x_1,x_2)=-2x_1^2+6x_1x_2-6x_1-5x_2^2+8x_2-5, \quad S=\R^2$
     \item $\ds f(x_1,x_2)=x_1^4-8x_1^3+24x_1^2-32x_1+4x_2^2-4x_2, \quad S=\R^2$
     \item $\ds f(x_1,x_2)=\sin(x_1)\cos(x_2),
	\quad S=\left\{(x_1,x_2) \mid 0 < x_1,x_2 < \pi \right\}$
     \item $\ds f(x_1,x_2)=\ln(1+x_1^2)+\frac{1}{2} x_2^2,
	\quad S=\left\{(x_1,x_2) \mid -1 < x_1 < 1 \right\}$
     \item $\ds f(x_1,x_2)=e^{-\left(\frac{x_1^2+x_2^2}{2}\right)},\quad S=\left\{(x_1,x_2) \mid x_1^2+x_2^2 <1 \right\}$ %global max; concave on domain; not hard to show;
    \ee
\item Suppose that $f:\R^n \to \R$ is the general quadratic mapping defined by
\begin{equation}\label{E:Exercises_sufficient_SECOND_order_convexity_quadratic}
f(\vecx)=\frac{1}{2}\vecx^tA\vecx-\vecb^t\vecx+c,
\end{equation}
where $A$ is a nonsingular, $n$-by-$n$ matrix, $\vecb$ is in $\R^n$, and $c$ is a real number.
Show that $\vecx=A^{-1}\vecb$ is the sole critical point of
$f$ and that this critical point is a global minimum (resp. global maximum) if $A$ is positive definite (resp. negative definite).
Use this result to determine the global minimum of $f$ in (\ref{E:Exercises_sufficient_SECOND_order_convexity_quadratic})
when $\ds A= \begin{bmatrix} 2 & 1 \\ 1 & 3 \end{bmatrix}$ and $\vecb= \begin{bmatrix} -1 \\ 4 \end{bmatrix}$, and $c=6$.
	
\item The general equation of a plane in $\R^3$ is given by $\ds \vecn^t \vecx=d$, where
$\vecn$ is a fixed vector in $\R^3$ (the \emph{normal vector}),
$\ds \vecx= \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}$, and $d$ is a constant.
Determine a formula in terms of $\vecn$ for the point, $\vecx_0$, on this plane that
is closest to the origin. Then find the point on the plane $\ds x_1+2x_2+3x_3=1$ that is closest to the origin.

\item Show that the function $f:\R^2 \to \R$ given by $f(x_1,x_2)=x_1^4+x_2^2$ has a critical point
at the origin, which is a global minimum. Then verify $ \ds H_f(\zerovec)$ is positive semidefinite.
Then show that $f(x_1,x_2)=-x_1^4+x_2^2$ has a critical point
at the origin, which is a saddle point, and verify that $ \ds H_f(\zerovec)$ is still positive semidefinite. This example illustrates how, in the hypothesis of part (2)
of Theorem \ref{Th:optimality_sufficient_SOC_Hessian_local_optimal_criteria}, we may not
replace ``positive definite'' with ``positive semidefinite.''

\item Give an example of a twice-differentiable function $f:\R^2 \to \R$ having
a global minimum at $\vecx_0=\zerovec$ and for which zero is the only eigenvalue of $\ds H_f(\vecx_0)$.
Then give an example of a twice-differentiable function \newline
$f:\R^2 \to \R$ having
a global maximum at $\vecx_0=\zerovec$ and for which zero is the only eigenvalue of $\ds H_f(\vecx_0)$.
%pm x^4+y^4%

\item Consider the zero-sum
involving the 2-by-2 payoff matrix
\begin{equation}\label{E:duality_payoff_matrix}
A=\begin{bmatrix} 2 & -3 \\ -1 & 4  \end{bmatrix}.
\end{equation}
Suppose player 1 chooses columns one and two with respective probabilities, $x_1$ and $x_2$. Player 2 chooses
rows one and two with respective probabilities $y_1$ and $y_2$.
\be
\item Formulate the LP and dual LP, whose respective
solutions indicate each player's optimal mixed strategy. Solve each of these LPs and determine the corresponding game value.
\item Confirm that the solution just obtained coincides with the saddle point of the function
\begin{equation*}
f(x_1,y_1)=\begin{bmatrix} y_1 \\ 1-y_1\end{bmatrix}^t A
\begin{bmatrix} x_1 \\ 1-x_1 \end{bmatrix}.
\end{equation*}
\item Sketch the set of points in the $x_1y_1$ plane that determine all possible mixed strategies for the two players that yield the game value from (a).
\item Now sketch the set of points in the $x_1y_1$ plane that determine all possible mixed strategies for the two players that yield a game value that is 20\% higher than that in (c).
\ee

\item Consider the zero-sum
having the 2-by-2 payoff matrix
\begin{equation}\label{E:duality_payoff_matrix}
A=\begin{bmatrix} a & b \\ c & d  \end{bmatrix}.
\end{equation}
Assume $a+d-(b+c) \ne 0$. Show that for the game to be fair, i.e., to have a game value of zero,
the matrix $A$ must be noninvertible.

\ee



