anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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anonymous
  • anonymous
what are the equations?
anonymous
  • anonymous
what is the function more importantly.
anonymous
  • 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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anonymous
  • anonymous
Maximize f(x,y,z) subject to the restriction g and h?
anonymous
  • anonymous
actually i know how to solve
anonymous
  • anonymous
but are these g and h related or both different
anonymous
  • anonymous
yeah maximize with restriction g and h i think
anonymous
  • anonymous
I would let G(x,y,z) = ( x^2 + y^2 , 2y + z) where this our constrained set.
anonymous
  • anonymous
so now partial derivative of f(x)=lamba(g(x))
anonymous
  • anonymous
yes,
anonymous
  • anonymous
\[\frac{ part( f}{par( x,y,z)} = (\lambda_1, \lambda_2, \lambda_3)\frac{par G}{par (x,y,z)}\]
anonymous
  • anonymous
if its two dimensional we need to take only 1 lambda then why 3 for 3D?
anonymous
  • 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
  • anonymous
I don't know how you need 1 eigenvalue for 2. You need 2.
anonymous
  • anonymous
lambda was probably written as a vector.
anonymous
  • anonymous
and it had two components.
anonymous
  • 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
  • anonymous
but here its 3D and i dont know how to solve this
anonymous
  • anonymous
Same business here.
anonymous
  • anonymous
your part(g)/part(x,yz) is going to be a matrix 3x3.
anonymous
  • anonymous
find its eigenvalues using characteristic polynomial
anonymous
  • anonymous
or however you find eigenvalues.
anonymous
  • 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
  • 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
  • anonymous
W(x,y,z) = ( x^2 + y^2 , 2y + z)
anonymous
  • anonymous
| 2x 2y 0 | | 0 2y 1 | | |
anonymous
  • 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
  • anonymous
even a derivative is technically a matrix, so you were right when you said you only needed one lambda
anonymous
  • 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
  • anonymous
ok i will try to solve and wil you be available online??
anonymous
  • anonymous
what do you get as partial f(x,y,z)
anonymous
  • anonymous
<1,0,2z>
anonymous
  • anonymous
<1,0,2z> = | 2x 2y 0 | | 0 2y 1 |
anonymous
  • 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
  • anonymous
here 3 equations and 5 unknowns
anonymous
  • anonymous
Don't worry we are just trying to find expressions for x,y,z not explicit answers. what did you get?
anonymous
  • anonymous
i got lam1=1/2x 2y(lam1+lam2)=0 lam2=2z
anonymous
  • anonymous
try to get expressions for x, y,z and remember you can still use you're functions g, h
anonymous
  • anonymous
BTW, are you sure the functions weren't equal to anything?
anonymous
  • anonymous
no
anonymous
  • anonymous
g(x,y,z)=x^2+y^2 h(x,y,z)=2y+z These two can describe tons of functions.
anonymous
  • 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
  • anonymous
if i solve i got 2 points (1,0,0) and (-1,0,0) are these correct??
anonymous
  • 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
  • anonymous
i solved the above equations with g(x) ,h(x)
anonymous
  • anonymous
x^2 + y^2 = 1 and 2y+z = 0 with the equations with lambdas
anonymous
  • anonymous
oh so they were equal to something.
anonymous
  • anonymous
COOL! i was getting scared.
anonymous
  • anonymous
am i doing correct or the points i got correct and the functional values at this point gives the value??
anonymous
  • anonymous
I hope so, don't make me do the algebra =P
anonymous
  • anonymous
x^2 + y^2 = 1 and 2y+z = 0 was this given?
anonymous
  • anonymous
yes

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