\input epsf
\nopagenumbers
\magnification=\magstep1

\noindent {\bf Problem statement}  The $2\times 2$ determinant can be thought of as a function which 
takes four variables as input, and returns a real number as output:
 
 $$\det(a,b,c,d)=\det\pmatrix{a&b\cr c&d\cr} = ad -bc $$
 
\noindent a) What is the gradient of this function, $\nabla \det$?
(The gradient of any function is a vector.  First question: how many
components will $\nabla\det$ have?)

\medskip

\noindent b) If $a=2$, $b=-3$, $c=4$, and $d=5$, then 
 
$$\det(a,b,c,d)=\det(2,-3,4,5)=\det\pmatrix{2&-3\cr 4&5\cr} 
  = 22. $$
 
\noindent Suppose we want to change each of $a$, $b$, $c$, and $d$ by
a little bit, where ``little bit'' here means that $(\triangle a)^2 +
(\triangle b)^2 + (\triangle c)^2 + (\triangle d)^2 \le .01$.  If we
want to make changes so the new determinant is as {\it large}\/ as
possible, what changes would you recommend?


\vfil\eject\end