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Math Help - Multivariable max&min

  1. #1
    Member Miss's Avatar
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    Multivariable max&min

    Find the max&min of the following function :
    f(x,y,z)=8xy-3y^2+32z+5
    subject to the constraint : 4x^2+y^2+4z^2=24

    By the Lagrang Multipliers' method :

    8y=\lambda 8x........(1)
    8x-6y=\lambda 2y.......(2)
    32=\lambda 8z.......(3)
    4x^2+y^2+4z^2=24

    (1) will be : y=\lambda x
    (2) will be : 4x-3y=\lambda y
    (3) will be : 4=\lambda z

    then ?!!
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  2. #2
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by Miss View Post
    Find the max&min of the following function :
    f(x,y,z)=8xy-3y^2+32z+5
    subject to the constraint : 4x^2+y^2+4z^2=24

    By the Lagrang Multipliers' method :

    8y=\lambda 8x........(1)
    8x-6y=\lambda 2y.......(2)
    32=\lambda 8z.......(3)
    4x^2+y^2+4z^2=24

    (1) will be : y=\lambda x
    (2) will be : 4x-3y=\lambda y
    (3) will be : 4=\lambda z

    then ?!!

    sub  y = \lambda x

    into

    4x- 3y = \lambda y

    and get

    -\lambda^2x+4x-3\lambda x = 0

    factor out and x

    x(-\lambda^2+4-3\lambda) = 0

    x= 0

    Now plug that x = 0 into

    y = \lambda x

    and get that y = 0

    Last plug x = y = 0

    into

    4x^2 +y^2 +4z^2 =24
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  3. #3
    Member Miss's Avatar
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    thanks !!

    But why did you ignore solving for \lambda in :

    x(-\lambda^2+4-3\lambda) = 0

    ??
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  4. #4
    Super Member General's Avatar
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    By using the method of Lagrange, we will get:

    y=\lambda x ... (1)
    4x-3y=\lambda y ... (2)
    4=\lambda z ... (3)
    4x^2+y^2+4z^2=24 ... (4)

    By substituting (1) in (2), we will get the equation : x(-\lambda^2 - 3\lambda + 4)=0
    \implies x=0,\lambda=-4,\lambda=1 ..

    when x=0 ---> y=0 , substitute this in 4 ---> z^2=6 \implies z=\pm \sqrt{6}
    so the first two points are (0,0,\sqrt{6}) and (0,0,-\sqrt{6})
    when \lambda = 1 ---> z=4 from (3) ---> y=x from (1) and (2)
    So when \lambda=1 we have y=x and z=4 , By substituting this in (4) :
    5x^2+64=24 \implies x^2=-8 which don't have any solutions
    so when \lambda = 1, we do not have any points.
    when \lambda=-4 ---> y=-4x from (1) and z=-1 from (3)
    by substituting this in (4) ---> x^2=1 \implies x=\pm 1
    when x=1 --> y=-4 ---> third point is (1,-4,-1)
    when x=-1 ---> y=4 ---> fourth point is (-1,4,-1)
    The points are : (0,0,\sqrt{6}),(0,0,-\sqrt{6}),(1,-4,-1) and (-1,4,-1)
    Find the value of the function at these point, biggest --> max, smallest --> min.
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  5. #5
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by Miss View Post
    thanks !!

    But why did you ignore solving for \lambda in :

    x(-\lambda^2+4-3\lambda) = 0

    ??
    Sorry forgot to copy that part from my worksheet when transfering it on here lol.
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