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{
\bf Dr. Z's Math251 Handout \#14.8 [Lagrange Multipliers]
}

By Doron Zeilberger

{\bf Problem Type 14.8a}: 
Use Lagrange multipliers
to find the maximum and minimum values of the
function subject to the given conditions.
$$
f(x,y,z)=Expression_(x,y,z) \quad ; \quad g(x,y,z)=k \quad .
$$
{\bf Example Problem 14.8a}: 
Use Lagrange multipliers
to find the maximum and minimum values of the
function subject to the given conditions.
$$
f(x,y,z)=xyz \quad ; \quad 3x^2+2y^2+z^2=6 \quad .
$$

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{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} Find the gradients of $f(x,y,z)$
and $g(x,y,z)$.

&
{\bf 1.} 
$f_x={{\partial} \over {\partial x}} xyz=yz$,
$f_y={{\partial} \over {\partial y}} xyz=xz$,
$f_z={{\partial} \over {\partial z}} xyz=xy$.
So,
$$
\g f = \langle yz,xz,xy \rangle \quad .
$$
$g_x={{\partial} \over {\partial x}} (3x^2+2y^2+z^2)=6x$,
$g_y={{\partial} \over {\partial y}} (3x^2+2y^2+z^2)=4y$,
$g_z={{\partial} \over {\partial z}} (3x^2+2y^2+z^2)=2z$.
So
$$
\g g = \langle 6x,4y,2z \rangle \quad.
$$
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{\bf 2.} Introduce another variable, $\lambda$, and set
up the equations implied by
$$
\g f= \lambda \g g \quad.
$$
To them add, the equation $g(x,y,z)=k$.
&
{\bf 2.} 
$\g f= \lambda \g g$ means
$$
\langle yz,xz,xy \rangle=\lambda \langle 6x,4y,2z \rangle \quad,
$$
that spells-out to the set of equations
$$
yz=6\lambda x \quad, \quad xz= 4 \lambda y \quad , 
$$
$$
\quad xy=2 \lambda z \quad, \quad 3x^2+2y^2+z^2=6 \quad .
$$

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{\bf 3.} Use algebra to solve the system of four equations
and four unknowns.

&
{\bf 3.} Multiplying the first three equations, we get
$(xyz)^2=48\lambda^3xyz$ so $xyz=48 \lambda^3$, and hence
$yz=48\lambda^3/x$ and we get from the first equation
$48\lambda^3/x=6 \lambda x$. This means
$8 \lambda^2=x^2$ and so $x=\sqrt{8} \lambda$.

From $xyz=48 \lambda^3$, we also get
$xz=48\lambda^3/y$ and we get from the second equation
$48\lambda^3/y=4 \lambda y$. This means
$12 \lambda^2=y^2$ and so $y=\sqrt{12} \lambda$.

From $xyz=48 \lambda^3$, we also get
$xy=48\lambda^3/z$ and we get from the third equation
$48\lambda^3/z=2 \lambda z$. This means
$24 \lambda^2=z^2$ and so $z=\sqrt{24} \lambda$.

Plugging these expressions in $\lambda$ into the last equation
$3x^2+2y^2+z^2=6$ we get
$$
3(\sqrt{8} \lambda)^2+
2( \sqrt{12} \lambda)^2+
(\sqrt{24} \lambda)^2=6 \quad .
$$
So,
$$
(3 \cdot 8+ 2 \cdot 12 + 24) \lambda^2=6 \quad ,
$$
$$
\lambda^2= {{1} \over {12}} \quad ,
$$
and so
$$
\lambda= \pm {{1} \over {\sqrt{12}}} \quad .
$$
We get {\bf two} solutions. The first one is
$$
\lambda={{1} \over {\sqrt{12}}} \quad, \quad
x=\sqrt{8} \cdot {{1} \over {\sqrt{12}}}=\sqrt{2/3} \quad,
$$
$$
y=\sqrt{12} \cdot {{1} \over {\sqrt{12}}}=1 \quad,\quad
z=\sqrt{24} \cdot {{1} \over {\sqrt{12}}}=\sqrt{2} \quad,
$$
which means the point $(\sqrt{2/3},1, \sqrt{2})$.

And the second is
$$
\lambda={{-1} \over {\sqrt{12}}} \quad, \quad
x=\sqrt{8} \cdot {{-1} \over {\sqrt{12}}}=-\sqrt{2/3} \quad,
$$
$$
y=\sqrt{12} \cdot {{-1} \over {\sqrt{12}}}=-1 \quad,
z=\sqrt{24} \cdot {{-1} \over {\sqrt{12}}}=-\sqrt{2} \quad,
$$
which means the point $(-\sqrt{2/3},-1, -\sqrt{2})$.

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{\bf 4.} Now you can forget about the $\lambda$ and plug-in
these point(s) into $f$ and see who gives the
largest value, that is the {\bf maximum value} and
who is the smallest, that is the {\bf minimum value}.
&
{\bf 4.} 
$$
f(\sqrt{2/3},1, \sqrt{2})=2/\sqrt{3} \quad,
$$
$$
f(-\sqrt{2/3},-1, -\sqrt{2})=-2/\sqrt{3} \quad .
$$
{\bf Ans.:} The maximum value is $2/\sqrt{3}$ and
the minimum value is $-2/\sqrt{3}$.


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