Question

Use the Lagrange multiplier technique to find the maximum and minimum values of on the ellipse defined by the equations and , and where they occur

Answer 1

what are the equations?

Answer 2

what is the function more importantly.

Answer 3

f(x,y,z)=x+z^2
g(x,y,z)=x^2+y^2
h(x,y,z)=2y+z

Answer 4

Maximize f(x,y,z) subject to the restriction g and h?

Answer 5

actually i know how to solve

Answer 6

but are these g and h related or both different

Answer 7

yeah maximize with restriction g and h i think

Answer 8

I would let G(x,y,z) = ( x^2 + y^2 , 2y + z)
where this our constrained set.

Answer 9

so now partial derivative of f(x)=lamba(g(x))

Answer 10

yes,

Answer 11

\[\frac{ part( f}{par( x,y,z)} = (\lambda_1, \lambda_2, \lambda_3)\frac{par G}{par (x,y,z)}\]

Answer 12

if its two dimensional we need to take only 1 lambda then why 3 for 3D?

Answer 13

I don't know rigorous you're calc 3 is. Do you need to prove that G is a compact set? and f is continuous, then by Extreme Value Theorem our max actually exists.

Answer 14

I don't know how you need 1 eigenvalue for 2. You need 2.

Answer 15

lambda was probably written as a vector.

Answer 16

and it had two components.

Answer 17

basically if you have 2D then
partial f(x,y)=partial (lambda (g,y))
then slove for lambda
and find x and y then we can get max or min

Answer 18

but here its 3D and i dont know how to solve this

Answer 19

Same business here.

Answer 20

your part(g)/part(x,yz) is going to be a matrix 3x3.

Answer 21

find its eigenvalues using characteristic polynomial

Answer 22

or however you find eigenvalues.

Answer 23

but here f(x,y,z)=lambda g(x,y,z)or f(x,y,z)=lambdah(x,y,z)
how i beed to proceed

Answer 24

Yea so what we did was COMBINE our restrictions into one function lets say W(f(x,y,z), h(x,y,z)) Thats what I meant by capital G up there.

Answer 25

W(x,y,z) = ( x^2 + y^2 , 2y + z)

Answer 26

| 2x 2y 0 |
| 0 2y 1 |
| |

Answer 27

turns out to be our Part(W)/dx,y,z.
We only need 2 lambdas in this case.
The eigenvalues depend on the rank of our matrix.

Answer 28

even a derivative is technically a matrix, so you were right when you said you only needed one lambda

Answer 29

so we have par f(x,y,z) = (lambda1, lambda2) | 2x 2y 0 |
| 0 2y 1 |
is your linear algebra good enough to find lambdas?

Answer 30

ok i will try to solve and wil you be available online??

Answer 31

what do you get as partial f(x,y,z)

Answer 32

<1,0,2z>

Answer 33

<1,0,2z> = <lam1, lam2> | 2x 2y 0 |
| 0 2y 1 |

Answer 34

multiply the lambda through the matrix.
then you should be able to pick out a linear system of equations.
Use that to find lam1 and lam2

Answer 35

here 3 equations and 5 unknowns

Answer 36

Don't worry we are just trying to find expressions for x,y,z not explicit answers.
what did you get?

Answer 37

i got lam1=1/2x
2y(lam1+lam2)=0
lam2=2z

Answer 38

try to get expressions for x, y,z and remember you can still use you're functions g, h

Answer 39

BTW, are you sure the functions weren't equal to anything?

Answer 40

no

Answer 41

g(x,y,z)=x^2+y^2
h(x,y,z)=2y+z
These two can describe tons of functions.

Answer 42

x^2 + y^2 = 5, is certainly not the same as x^2 + y^2 = 10, and they will have max's in different points.

Answer 43

if i solve i got 2 points (1,0,0) and (-1,0,0)
are these correct??

Answer 44

or are we doing a general case? set them equal to x^2 + y^2 = a
2y+z = b.

How did you do that..

Answer 45

i solved the above equations with g(x) ,h(x)

Answer 46

x^2 + y^2 = 1 and 2y+z = 0 with the equations with lambdas

Answer 47

oh so they were equal to something.

Answer 48

COOL! i was getting scared.

Answer 49

am i doing correct or the points i got correct and the functional values at this point gives the value??

Answer 50

I hope so, don't make me do the algebra =P

Answer 51

x^2 + y^2 = 1 and 2y+z = 0 was this given?

Answer 52

yes