# Thread: global max an global min

1. ## global max an global min

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?

2. Hello, yzc717;230977!

I would use Lagrange Multipliers on #4 . . .

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

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

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

Take partial derivatives and equate to zero . . .

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

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

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

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

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

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