Question

use langrange multipliers to find the extrema of f subject to the stated constraints.

f(xyz) = x + y + z
constraint = x^2+y^2+z^2=25

Answer 1

this is what i've gotten so far and i'm not sure if i'm doing it right / where to go from here...
\[x+y+z - \lambda (x^2+y^2+z^2-25) \]
\[x+y+z-\lambda x^2-\lambda y^2-\lambda z^2 +25\lambda\]
\[Fx = 1-2x \lambda \]
\[fy= 1 - 2y \lambda\]
\[fz= 1 - 2z \lambda\]
\[f \lambda -x^2 - y^2-z^2 +25\]

\[x = 1/2\lambda\]
\[y = 1/2\lambda\]
\[z=1/2\lambda\]
\[\lambda = 1/33.3\]

Answer 2

please halp ._.

Answer 3

\[ \nabla f(x,y,z) = \lambda \nabla(x^2 + y^2 + z^2)\]

Answer 4

yeah, that's what i did to get x+y+z−λx2−λy2−λz2+25λ

Answer 5

\[ <1, 1, 1> = \lambda <2x, 2y,2z> \]

Answer 6

wait.. why is it < 1 1 1 > ??

Answer 7

nvm, its the derivates.. you're taking the derivatives first

Answer 8

\( \nabla \) is gradient operator

Answer 9

my book / professor has us do the equation out first, and then separately find fx, fy, fz, and flambda

is there a reason why you've done it backwards? does this make it easier? how do i find the extrema with this?

Answer 10

I am not so sure ... i've just learned a day back.
http://ocw.mit.edu/courses/mathematics/18-02-multivariable-calculus-fall-2007/video-lectures/lecture-13-lagrange-multipliers/
If you try to visualise at maximum or minimum, the normal to curves are parallel to each other .... it makes much sense. They say so .... and i agree!!

Answer 11

In this case, it is minimization of function!!

Answer 12

how do you know that? :o and how do i do that?

Answer 13

picture ... of the function and constraint.
a plane and a sphere (i guess both maximum and minimum value are same ... in this case ... because of symmetricity)

Answer 14

i'm not sure i follow but.. the answer is
min f(-5/root3, -5/root3, -5/root3) = -5/root3
max: same but everything is positive..
is my value of lambda incorrect? =\

Answer 15

nvms i gots it ^.^ thanks!