MATH 251 Lecture 37

« previous | "Friday", May 1, 2012 | next »

Review Session

Lagrange Multipliers

$f(x,y,z)=xyz$, constraint: $g(x)=x^{2}+2y^{2}+3z^{2}=6$

Find extrema of $f$ given $g=6$.

${\begin{aligned}{\vec {\nabla }}f&=\lambda {\vec {\nabla }}g\\yz&=\lambda 2x\\xz&=\lambda 4y\\xy&=\lambda 6z\\x^{2}+2y^{2}+3z^{2}&=6\end{aligned}}$

Solving the system of equations gives

${\begin{aligned}x&=\pm {\sqrt {2}}\\y&=\pm 1\\z&=\pm {\sqrt {\tfrac {2}{3}}}\end{aligned}}$

for a total of 8 possible points.

($\lambda$ is meaningful only if $Q(x,y)=ax^{2}+2bxy+cy^{2}$ is a quadric surface and $g(x,y)=x^{2}+y^{2}=1$ is the unit circle)

Finding Extrema in a region

Find the extrema of $f(x,y)=2x^{2}+x+y^{2}-2$ within $D=\{x^{2}+y^{2}=4\}$.

1. look interior: ${\vec {\nabla }}f=0$
2. look on boundary: $f(x,y)=x^{2}+x+2$ (substitute $x^{2}+y^{2}=4$)

Solving 1 gives us $(-{\tfrac {1}{4}},0)$ easily.

Solve for 2:

${\begin{aligned}x&=2\cos {\theta }\\y&=2\sin {\theta }\\f(\theta )&=8\cos ^{2}{\theta }+2\cos {\theta }+4\sin ^{2}{\theta }-2\\f'(\theta )&=(2\sin {\theta })(4\cos {\theta }+1)=0\\\theta &=0,\pi \\\cos {\theta }&=-{\tfrac {1}{4}}\end{aligned}}$

We now have 5 points to check: (-2,0), (2,0), (-1/2, ±sqrt(15)/2), and (-1/4,0)