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\centerline
{
\bf Solutions to the ``QUIZ'' for Lecture 10
}

{\bf 1.} 
Find the local maximum and minimum point(s),
the local maximum and minimum values,
and saddle point(s) of the function 
$$
f(x,y)=12x^2-4x^3+6y^2+12xy \quad .
$$

{\bf Sol.} First compute the first partial derivatives:
$$
f_x=24x-12x^2+12y \quad , \quad f_y=12y+12x \quad ;
$$
and for future reference, the second-order (partial) derivaties:
$$
f_{xx}=24-24x \quad , \quad f_{xy}=12 \quad , \quad f_{yy}=12\quad .
$$

To find the {\bf critical points}, we set both $f_x$ and $f_y$
to $0$ and
solve the system of two equations and two unknwons ($x$ and $y$).
$$
24x-12x^2+12y=0 \quad , \quad 12y+12x=0 \quad .
$$
Since the second one is simpler, let's treat it first, getting
$y=-x$. Plugging-in $y=-x$ into the first equation gives:
$$
24x-12x^2+12(-x)=0 \quad
$$
that simplifies to
$$
12x-12x^2=0 \quad .
$$
Factoring, we get
$$
12x(1-x)=0 \quad,
$$
that yields {\bf two} solutions $x=0$ and $x=1$ for the $x$-coordinate.
Since $y=-x$, if $x=0$ then $y=-0=0$ yielding the point $(0,0)$,
and if $x=1$, $y=-1$, yielding the point $(1,-1)$.
So the critical points are $(0,0)$ and $(1,-1)$.

For each of them, we have to determine whether they are local max., local
min. or saddle points.

For the candidate point $(0,0)$, plugging-in $x=0$, $y=0$ into
$f_{xx},f_{xy},f_{yy}$ yields:
$$
f_{xx}=24 \quad , \quad f_{xy}=12 \quad , \quad f_{yy}=12\quad .
$$
The {\bf discriminant} $D=f_{xx}f_{yy}-f_{xy}^2$ equals, in this case,
$24 \cdot 12 - 12^2 =144$, and since $f_{xx} >0$ it follows that
$(0,0)$ is a {\bf local min} (remember that if $f_{xx}>0$ then it
is a local min, while of $f_{xx}<0$ then it is a local max.)
The {\bf local min. value} is $f(0,0)=0$.

Now let's investigate the other candidate point $(1,-1)$. For that point
$$
f_{xx}=0 \quad , \quad f_{xy}=12 \quad , \quad f_{yy}=12\quad ,
$$
and $D=0 \cdot 12 - 12^2 =-144$ . Since $D<0$, it follows that
$(1,-1)$ is a {\bf saddle point}.

{\bf Ans.}: Local max: None;
Local min value is $0$ located at the local min point $(0,0)$;
Saddle point: $(1,-1)$.

{\bf Comments}: Unfortunately, only about \%30 of the students
got it completely correct, but many students followed the method correctly,
but sooner or later messed up the algebra. Some people ``almost'' got
it perfectly, but said that $(0,0)$ is a local max rather than the correct
answer that it is a local min. Remember that if $f_{xx}$ is {\bf positive} then
it is a local {\bf min} and vice versa.

Some people need to brush up on their algebra skills.
Some people solved $12x(1-x)=0$ and only got the solution $x=1$.
Make sure you find {\bf all} solutions.



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