MATH 251 Lecture 37

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Review Session

Lagrange Multipliers

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

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

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

Solving the system of equations gives

$\begin{array}{rl}x& =\pm \sqrt{2}\\ y& =\pm 1\\ z& =\pm \sqrt{\frac{2}{3}}\end{array}$

for a total of 8 possible points.

($\lambda$ is meaningful only if $Q(x,y)=a{x}^2+2bxy+c{y}^2$ 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: $\overrightarrow{\mathrm{\nabla }}f=0$
2. look on boundary: $f(x,y)=x^2+x+2$ (substitute $x^2+y^2=4$)

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

Solve for 2:

$\begin{array}{rl}x& =2\cos \theta \\ y& =2\sin \theta \\ f(\theta )& =8{\cos }^2\theta +2\cos \theta +4{\sin }^{}\theta -2\\ f^{\prime }(\theta )& =(2\sin \theta )(4\cos \theta +1)=0\\ \theta & =0,\pi \\ \cos \theta & =-\frac{1}{4}\end{array}$

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