global max an global min

• Dec 3rd 2008, 10:27 AM
yzc717
global max an global min
http://farm4.static.flickr.com/3293/...6bc5d8bd_o.png

my try on problem2 was take the partial devaritive eaqual to zero, get the C.P. then take the boundary point (3,2) , to find the max and min. it that right?

also, how to do problem#4?
• Dec 3rd 2008, 03:45 PM
Soroban
Hello, yzc717;230977!

I would use Lagrange Multipliers on #4 . . .

Quote:

4. Find the points on the curve of intersection of the surfaces: . $\begin{array}{c}x^2 - xy + y^2 - z^2 \:=\:1 \\ x^2+y^2+1 \end{array}$
in $\Re^2$ that are closest to the origin.

We want to minimize: . $d(x,y,z) \:=\:x^2+y^2+z^2$
. with the constraints: . $x^2-xy+y^2-z^2 -1\:=\:0\:\text{ and }\;x^2+y^2-1\:=\:0$

We have: . $f(x,y,z,\lambda,\mu) \;=\;x^2+y^2+z^2 + \lambda(x^2-xy+y^2-z^2-1) + \mu(x^2+y^2-1)$

Take partial derivatives and equate to zero . . .

. . $f_x \;=\;2x + \lambda(2x-y) + \mu(2x) \;=\;0$

. . $f_y \;=\;2y + \lambda(-x+2y) + \mu(2y) \;=\;0$

. . $f_z \;=\;2z + \lambda(-2z) \;=\;0$

. . $f_{\lambda} \;=\;x^2-xy+y^2-z^2-1\;=\;0$

. . $f_{\mu} \;=\;x^2+y^2 -1\;=\;0$

Then solve the system for: . $\lambda, \mu, x, y, z.$