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\leftline{Mathematics 241 -- Multivariable Calculus}
\leftline{Solution for Quiz 7 -- November 9, 2007}

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(10) Find the critical points of the function $f(x,y) = 3 x + 2 y$ on the 
constraint set $g(x,y) = x^2 + y^2 - 4 = 0$ by the method of Lagrange multipliers.
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{\it Solution:}  The Lagrange multiplier equations are
$$\eqalign{
{\partial f\over\partial x} &= \lambda {\partial g\over\partial x}\cr
{\partial f\over\partial y} &= \lambda {\partial g\over\partial y}\cr
g(x,y) &=0 \cr}$$
or
$$\eqalign{
3 &= 2 \lambda x\cr
2 &= 2 \lambda y\cr
x^2 + y^2 - 4  &=0. \cr}$$
From the first two equations,
$$3y = 2\lambda xy = 2x,$$
so we can eliminate $\lambda$ to obtain:
$$y = {2x\over 3}.$$
Substitute this into the constraint equation to eliminate $y$:
$$x^2 + \left({2x \over 3}\right)^2 = x^2 + {4x^2 \over 9} = {13 x^2 \over 9} = 4.$$
This says $x^2 = {36\over 13}$, so $x = \pm {6\over \sqrt{13}}$.  Then from 
the equation $y = {2x\over 3}$, we get the corresponding values of $y$, 
$y = \pm {4\over \sqrt{13}}$.  Note
that $y$ must have the same sign as $x$, so there are only {\it two} critical
points here:
$$\left({6\over \sqrt{13}}, {4\over \sqrt{13}}\right), \quad \left(-{6\over \sqrt{13}}, -{4\over \sqrt{13}}\right).$$
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