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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

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  1. anonymous
    • 5 years ago
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    what are the equations?

  2. anonymous
    • 5 years ago
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    what is the function more importantly.

  3. anonymous
    • 5 years ago
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    f(x,y,z)=x+z^2 g(x,y,z)=x^2+y^2 h(x,y,z)=2y+z

  4. anonymous
    • 5 years ago
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    Maximize f(x,y,z) subject to the restriction g and h?

  5. anonymous
    • 5 years ago
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    actually i know how to solve

  6. anonymous
    • 5 years ago
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    but are these g and h related or both different

  7. anonymous
    • 5 years ago
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    yeah maximize with restriction g and h i think

  8. anonymous
    • 5 years ago
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    I would let G(x,y,z) = ( x^2 + y^2 , 2y + z) where this our constrained set.

  9. anonymous
    • 5 years ago
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    so now partial derivative of f(x)=lamba(g(x))

  10. anonymous
    • 5 years ago
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    yes,

  11. anonymous
    • 5 years ago
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    \[\frac{ part( f}{par( x,y,z)} = (\lambda_1, \lambda_2, \lambda_3)\frac{par G}{par (x,y,z)}\]

  12. anonymous
    • 5 years ago
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    if its two dimensional we need to take only 1 lambda then why 3 for 3D?

  13. anonymous
    • 5 years ago
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    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.

  14. anonymous
    • 5 years ago
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    I don't know how you need 1 eigenvalue for 2. You need 2.

  15. anonymous
    • 5 years ago
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    lambda was probably written as a vector.

  16. anonymous
    • 5 years ago
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    and it had two components.

  17. anonymous
    • 5 years ago
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    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

  18. anonymous
    • 5 years ago
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    but here its 3D and i dont know how to solve this

  19. anonymous
    • 5 years ago
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    Same business here.

  20. anonymous
    • 5 years ago
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    your part(g)/part(x,yz) is going to be a matrix 3x3.

  21. anonymous
    • 5 years ago
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    find its eigenvalues using characteristic polynomial

  22. anonymous
    • 5 years ago
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    or however you find eigenvalues.

  23. anonymous
    • 5 years ago
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    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

  24. anonymous
    • 5 years ago
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    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.

  25. anonymous
    • 5 years ago
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    W(x,y,z) = ( x^2 + y^2 , 2y + z)

  26. anonymous
    • 5 years ago
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    | 2x 2y 0 | | 0 2y 1 | | |

  27. anonymous
    • 5 years ago
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    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.

  28. anonymous
    • 5 years ago
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    even a derivative is technically a matrix, so you were right when you said you only needed one lambda

  29. anonymous
    • 5 years ago
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    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?

  30. anonymous
    • 5 years ago
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    ok i will try to solve and wil you be available online??

  31. anonymous
    • 5 years ago
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    what do you get as partial f(x,y,z)

  32. anonymous
    • 5 years ago
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    <1,0,2z>

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

  34. anonymous
    • 5 years ago
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    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

  35. anonymous
    • 5 years ago
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    here 3 equations and 5 unknowns

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

  37. anonymous
    • 5 years ago
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    i got lam1=1/2x 2y(lam1+lam2)=0 lam2=2z

  38. anonymous
    • 5 years ago
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    try to get expressions for x, y,z and remember you can still use you're functions g, h

  39. anonymous
    • 5 years ago
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    BTW, are you sure the functions weren't equal to anything?

  40. anonymous
    • 5 years ago
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    no

  41. anonymous
    • 5 years ago
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    g(x,y,z)=x^2+y^2 h(x,y,z)=2y+z These two can describe tons of functions.

  42. anonymous
    • 5 years ago
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    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.

  43. anonymous
    • 5 years ago
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    if i solve i got 2 points (1,0,0) and (-1,0,0) are these correct??

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

  45. anonymous
    • 5 years ago
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    i solved the above equations with g(x) ,h(x)

  46. anonymous
    • 5 years ago
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    x^2 + y^2 = 1 and 2y+z = 0 with the equations with lambdas

  47. anonymous
    • 5 years ago
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    oh so they were equal to something.

  48. anonymous
    • 5 years ago
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    COOL! i was getting scared.

  49. anonymous
    • 5 years ago
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    am i doing correct or the points i got correct and the functional values at this point gives the value??

  50. anonymous
    • 5 years ago
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    I hope so, don't make me do the algebra =P

  51. anonymous
    • 5 years ago
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    x^2 + y^2 = 1 and 2y+z = 0 was this given?

  52. anonymous
    • 5 years ago
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    yes

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