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

- anonymous

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- anonymous

what are the equations?

- anonymous

what is the function more importantly.

- anonymous

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

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## More answers

- anonymous

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

- anonymous

actually i know how to solve

- anonymous

but are these g and h related or both different

- anonymous

yeah maximize with restriction g and h i think

- anonymous

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

- anonymous

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

- anonymous

yes,

- anonymous

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

- anonymous

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

- anonymous

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.

- anonymous

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

- anonymous

lambda was probably written as a vector.

- anonymous

and it had two components.

- anonymous

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

- anonymous

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

- anonymous

Same business here.

- anonymous

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

- anonymous

find its eigenvalues using characteristic polynomial

- anonymous

or however you find eigenvalues.

- anonymous

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

- anonymous

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.

- anonymous

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

- anonymous

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

- anonymous

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.

- anonymous

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

- anonymous

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?

- anonymous

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

- anonymous

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

- anonymous

<1,0,2z>

- anonymous

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

- anonymous

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

- anonymous

here 3 equations and 5 unknowns

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

no

- anonymous

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

- anonymous

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.

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

oh so they were equal to something.

- anonymous

COOL! i was getting scared.

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

yes

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