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anonymous
 5 years ago
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
 5 years ago
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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what are the equations?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what is the function more importantly.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0f(x,y,z)=x+z^2 g(x,y,z)=x^2+y^2 h(x,y,z)=2y+z

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Maximize f(x,y,z) subject to the restriction g and h?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0actually i know how to solve

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but are these g and h related or both different

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah maximize with restriction g and h i think

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I would let G(x,y,z) = ( x^2 + y^2 , 2y + z) where this our constrained set.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so now partial derivative of f(x)=lamba(g(x))

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ part( f}{par( x,y,z)} = (\lambda_1, \lambda_2, \lambda_3)\frac{par G}{par (x,y,z)}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if its two dimensional we need to take only 1 lambda then why 3 for 3D?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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
 5 years ago
Best ResponseYou've already chosen the best response.0I don't know how you need 1 eigenvalue for 2. You need 2.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lambda was probably written as a vector.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and it had two components.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0basically 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
 5 years ago
Best ResponseYou've already chosen the best response.0but here its 3D and i dont know how to solve this

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0your part(g)/part(x,yz) is going to be a matrix 3x3.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0find its eigenvalues using characteristic polynomial

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0or however you find eigenvalues.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but 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
 5 years ago
Best ResponseYou've already chosen the best response.0Yea 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
 5 years ago
Best ResponseYou've already chosen the best response.0W(x,y,z) = ( x^2 + y^2 , 2y + z)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0 2x 2y 0   0 2y 1   

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0turns 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
 5 years ago
Best ResponseYou've already chosen the best response.0even a derivative is technically a matrix, so you were right when you said you only needed one lambda

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so 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
 5 years ago
Best ResponseYou've already chosen the best response.0ok i will try to solve and wil you be available online??

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what do you get as partial f(x,y,z)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0<1,0,2z> = <lam1, lam2>  2x 2y 0   0 2y 1 

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0multiply 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
 5 years ago
Best ResponseYou've already chosen the best response.0here 3 equations and 5 unknowns

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Don't worry we are just trying to find expressions for x,y,z not explicit answers. what did you get?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i got lam1=1/2x 2y(lam1+lam2)=0 lam2=2z

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0try to get expressions for x, y,z and remember you can still use you're functions g, h

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0BTW, are you sure the functions weren't equal to anything?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0g(x,y,z)=x^2+y^2 h(x,y,z)=2y+z These two can describe tons of functions.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x^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
 5 years ago
Best ResponseYou've already chosen the best response.0if i solve i got 2 points (1,0,0) and (1,0,0) are these correct??

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0or are we doing a general case? set them equal to x^2 + y^2 = a 2y+z = b. How did you do that..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i solved the above equations with g(x) ,h(x)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x^2 + y^2 = 1 and 2y+z = 0 with the equations with lambdas

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh so they were equal to something.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0COOL! i was getting scared.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0am i doing correct or the points i got correct and the functional values at this point gives the value??

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I hope so, don't make me do the algebra =P

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x^2 + y^2 = 1 and 2y+z = 0 was this given?
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