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\begin{document}
\leftline{MATH 241}
\leftline{Solution for Quiz 7}
\leftline{October 11, 2013}
\vskip 10pt
\noindent
A) (20) Using the method of Lagrange Multipliers, find the maximum and minimum values of $f(x,y) = 2x + 3y$ on 
the constraint curve
$$E = \left\{(x,y) \mid x^2 + \frac{y^2}{9} = 1\right\}.$$
(If you run into trouble with the algebra, for more partial credit, sketch what the level curves of $f$ 
should look like at the maximum and minimum together with the constraint curve.)
\vskip 10pt
\noindent
{\it Solution:}  We have $f_x = 2$, $f_y = 3$ and $g_x = 2x$, $g_y = \frac{2y}{9}$.  Therefore the Lagrange
Equations are 
\begin{eqnarray*}
2 &=& 2\lambda x\\
3 &=& \frac{2\lambda y}{9}\\
x^2 + \frac{y^2}{9} &=& 1
\end{eqnarray*}
If we multiply the first equation by $y$ and the second by $9 x$, then we get
\begin{equation}
\label{No1}
2y = 2\lambda xy = 27x,
\end{equation}
so $y = \frac{27x}{2}$.  Substituting into the constraint equation, we get
$$x^2\left(1 + \frac{81}{4}\right) = 1 \Rightarrow x = \pm \frac{2}{\sqrt{85}}$$
Then equation (\ref{No1}) says 
$$y = \pm \frac{27}{\sqrt{85}}$$
Note equation (\ref{No1}) implies 
that the positive sign for $x$ gives a positive $y$ and the negative sign for $x$ gives
a negative $y$.  This means that there are \emph{only two critical points of $f$ restricted
to $E$}:
$$\left(\frac{2}{\sqrt{85}},\frac{27}{\sqrt{85}}\right), \left(\frac{-2}{\sqrt{85}},\frac{-27}{\sqrt{85}}\right)$$
The function $f(x,y) = 2x + 3y$ has a positive value at the first point and a negative value at the second
point, so the first gives the maximum value $\sqrt{85}$, and the second gives the minimum value $-\sqrt{85}$.  

The geometric meaning of the Lagrange equations is that the level curves for $f(x,y) = 2x + 3y$ at the 
maximum and minimum should be \emph{tangent to the ellipse} $E$.  The level curves of $f$ are the lines
$2x + 3y = c$, or $y = \frac{-2}{3}x + \frac{c}{3}$, so these are straight lines of slope $m = \frac{-2}{3}$. 
Here is a picture of the constraint curve and the two contours for the maximum and minimum values:

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