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\centerline
{
\bf Dr. Z's Math251 Handout \#14.7 (2nd ed.)
[Optimization in Several Variables]
}

By Doron Zeilberger

{\bf Problem Type 14.7a}: Find the local maximum and
minimum values and saddle point(s) of the function $f(x,y)$.

{\bf Example Problem 14.7a}: Find the local maximum and
minimum values and saddle point(s) of the function
$$
f(x,y)=x^3y+12x^2-8y \quad .
$$

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{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} Find the partial derivatives $f_x$ and $f_y$.
For future reference, also compute $f_{xx}$, $f_{xy}$ and
$f_{yy}$.

&
{\bf 1.} 
$$
f_x={{\partial} \over {\partial x}}(x^3y+12x^2-8y)=3x^2y+24x \quad ,
$$
$$
f_y={{\partial} \over {\partial y}}(x^3y+12x^2-8y)=x^3-8 \quad ,
$$
$$
f_{xx}={{\partial} \over {\partial x}}(3x^2y+24x)=6xy+24 \quad ,
$$
$$
f_{xy}={{\partial} \over {\partial y}}(3x^2y+24x)=3x^2 \quad ,
$$
$$
f_{yy}={{\partial} \over {\partial y}}(x^3-8)=0 \quad .
$$
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{\bf 2.} Set both $f_x$ and $f_y$ to zero. Then solve the
system of two equations and two unknowns. The
solutions are the {\bf critical points}.

&
{\bf 2.} 
$$
3x^2y+24x=0 \quad, \quad x^3-8=0 \quad .
$$
Which is the same as
$$
x(3xy+24)=0 \quad, \quad  x^3-8=0 \quad .
$$
From the second equation we get $x=2$, and plugging it
into the first we get \hfill\break
$2 \cdot (6y+24)=0$, so $y=-4$.
It turns out that in this problem there is only one critical point:
$(2,-4)$.
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{\bf 3.} Plug the point(s), one at a time, into
$f_{xx},f_{xy}, f_{zz}$ and for each compute
the {\bf discriminant} $D=f_{xx}f_{yy}-[f_{xy}]^2$.
At the examined point $(a,b)$:

If $D>0$ and $f_{xx}>0$ then $(a,b)$ is a {\bf local minimum}
and the local minimum value is $f(a,b)$.

If $D>0$ and $f_{xx}<0$ then $(a,b)$ is a {\bf local maximum}
and the local maximum value is $f(a,b)$.

If $D<0$, then $(a,b)$ is neither max. nor min
but a {\bf saddle point}.

If $D=0$ then we {\bf don't know} (the test is inconclusive).

&
{\bf 3.} 
$$
f_{xx}(2,-4)=6(2)(-4)+24=-24 \quad ,
$$
$$
f_{xy}(2,-4)=3\cdot 2^2=12 \quad ,
$$
$$
f_{yy}(2,-4)=0 \quad .
$$
Since $D=(-24) \cdot 0-12^2=-144$ is {\bf negative},
this is a {\bf saddle point}.

{\bf Ans.}: The function has no maximum values and no
mimimum values. It has one saddle point at $(2,-4)$.

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\vfill\eject

{\bf Problem Type 14.7b}: 
Find the absolute maximum and minimum values
of $f$ on the set $D$
$$
f(x,y)=Expression(x,y) \quad ,
$$
$$
S=\{(x,y) \, \vert \, a_1 \leq x \leq a_2
\, , \, b_1 \leq y \leq b_2 \, \} \quad .
$$

{\bf Example Problem 14.7b}: Find the absolute maximum and minimum values
of $f$ on the set $S$
$$
f(x,y)=4x+6y-x^2-y^2 \quad ,
$$
$$
S=\{(x,y) \, \vert \, 0 \leq x \leq 4 \, , \, 
0 \leq y \leq 5 \, \} \quad .
$$

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{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} First find the {\bf critical points}
by computing $f_x$, $f_y$, setting them both equal
to $0$, and solving for $x$ and $y$. Only
retain those points that belong to $S$.
Then plug-in these point(s) into $f$,
and keep them for comparison later on.

%%%End General Step 1
&
{\bf 1.} 
$f_x=4-2x$, $f_y=6-2y$.
Solving
$$
4-2x=0 \quad , \quad 6-2y=0 \quad ,
$$
gives one solution $x=2,y=3$. So $(2,3)$ is a critial point.
Also $f(2,3)=4 \cdot 2+6 \cdot 3-2^2-3^2=8+18-4-9=13$.
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{\bf 2.} For each part of the boundary (in the case of
a rectangle there are four sides), find the absolute
max and min, like you did way back in Calc I.

&
{\bf 2.} 
On the LEFT side $x=0$ and $0 \leq y \leq 5$.
$f(0,y)=6y-y^2$. Let's call this function, for now
$F(y)$.
$F'(y)=6-2y$ which is $0$ at $y=3$. 
$F(3)=6 \cdot 3 -3^2=18-9=9$.
At the endpoints $F(0)=0, F(5)=5$.
So looking at the numbers $0,5,9$, the largest is $9$
and the smallest is $0$

{\bf abs. min. on Left Side}: $0$,

{\bf abs. max. on Left Side}: $9$.


On the RIGHT side $x=4$ and $0 \leq y \leq 5$.
$f(4,y)=16+6y-16-y^2=6y-y^2$. Let's call this function, for now
$F(y)$.
$F'(y)=6-2y$ which is $0$ at $y=3$. 
$F(3)=6 \cdot 3 -3^2=18-9=9$.
At the endpoints $F(0)=0, F(5)=5$.
So looking at the numbers $0,5,9$, the largest is $9$
and the smallest is $0$

{\bf abs. min. on Right Side}: $0$,

{\bf abs. max. on Right Side}: $9$.


On the DOWN side $y=0$ and $0 \leq x \leq 4$.
$f(x,0)=4x-x^2$. Let's call this function, for now
$F(x)$.
$F'(x)=4-2x$ which is $0$ at $x=2$, and
$F(2)=9$.
At the endpoints $F(0)=0, F(4)=0$.
So looking at the numbers $4,0,0$, the largest is $4$
and the smallest is $0$

{\bf abs. min. on DOWN Side}: $0$,

{\bf abs. max. on DOWN Side}: $4$.

On the UP side $y=5$ and $0 \leq x \leq 4$.
$f(x,5)=4x-x^2+5$. Let's call this function, for now
$F(x)$.
$F'(x)=4-2x$ which is $0$ at $x=2$, and
$F(2)=9$.
At the endpoints $F(0)=5, F(4)=5$.
So looking at the numbers $4,5,9$, the largest is $9$
and the smallest is $4$

{\bf abs. min. on UP Side}: $4$,

{\bf abs. max. on UP Side}: $9$.



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{\bf 3.} Now gather all these champions (in both 
min. and max. catergories) plus those came from the
critical points inside the region and find the largest
value, this is your {\bf absolute maximum value}
and the smallest, this is your {\bf absolute minimum value}.


&
{\bf 3.} 
For abs. min the contdenders are
$0,0,0,4,13$ so the {\bf absolute minimum value} is $0$.

For abs. max the contdenders are
$9,9,4,5,13$ so the {\bf absolute maximum value} is $13$.



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{\bf Problem Type 14.7c}: 
Find the point on the surface
$F(x,y,z)=k$ that is closest to the origin.


{\bf Example Problem 14.7c}: Find the point on the surface
$x^2y^2z=1$ that is closest to the origin.


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{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} It is more convenient to consider
the distance-squared, which is
$x^2+y^2+z^2$. Take one of the variables (say $z$)
(whatever is convenient) and express it in terms
of the other two (say $x,y$), and plug it into $x^2+y^2+z^2$
getting a function, let's call it $f(x,y)$
&
{\bf 1.} $z=1/(x^2y^2)$ so $z^2=x^{-4}y^{-4}$
and the distance-squared, in terms of $x,y$ is
$f(x,y)=x^2+y^2+x^{-4}y^{-4}$.

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{\bf 2.} Find the critical points by taking
$f_x,f_y$ and setting them equal to zero, and solving
for $x$ and $y$.
&
{\bf 2.} 
$$
f_x= {{\partial} \over {\partial x}} (x^2+y^2+x^{-4}y^{-4})=
2x-4x^{-5}y^{-4} \quad ,
$$
$$
f_y= {{\partial} \over {\partial y}} (x^2+y^2+x^{-4}y^{-4})=
2y-4x^{-4}y^{-5} \quad .
$$
We have to solve
$$
2x-4x^{-5}y^{-4}=0 \quad , \quad 2y-4x^{-4}y^{-5}=0 \quad ,
$$
which is the same
$$
x^6y^4=2 \quad , \quad x^4y^6=2 \quad .
$$
Dividing the first by the second we get $x^2/y^2=1$ so
$x^2=y^2$ and $x^{10}=2$ and
we get $x^2=y^2=2^{1/5}$ so
$x=\pm 2^{1/10}$, $y=\pm 2^{1/10})$.


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{\bf 3.} To get the $z$ coordinates for each of these points
plug into $f(x,y)$.
&
{\bf 3.} 
$z=1/(x^2y^2)$ so for each of the four possibilities
$z=1/2^{2/5}=2^{-2/5}$.

{\bf Ans.:} The points on the surface closest to the origin are
$(\pm 2^{1/10},\pm 2^{1/10},2^{-2/5})$.


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{\bf Problem from a Previous Final}

Find the local maximum and minimum points, the local maximum
and minimum values, and saddle points of the function
$$
f(x,y)= 4x^2+y^2+2x^2y-1 \quad .
$$

{\bf Ans.}: $(0,0)$ is the local minimum point

$-1$ is the local minimum value

$(\sqrt 2, -2), (-\sqrt 2, -2)$ are saddle points.


{\bf Another Problem from a Previous Final}

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

{\bf Ans.} Local max: none. Local Min: location: $(0,0)$, value: $0$.
Saddle point: $(-1,1)$.

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