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Math Help - calc 3, maxes and mins. quick question

  1. #1
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    calc 3, maxes and mins. quick question

    find the maximum and minimum values of the function subject to the given contraint.
    f(x,y)= 8x+3y
    subject to (x-1)^2 + (y+2)^2 = 9

    so i took the gradient of each..
    (8, 3)
    (2x-2, 2y+2)

    and to solve for the max and min, we need to use lagrange multipliers. so the gradient of f is equal to the gradient of g times lambda (some constant)

    i have the equations
    8= lambda(2x-2)
    3=lambda(2y+4)
    (x-1)^2 + (y+2)^2 =9

    how do i solve this system of equations? the answer in the back of the book is kind of crazy...
    x= 1 - 24/sqrt 73
    y= -2- 9/sqrt 73
    lambda= -sqrt 73/6
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  2. #2
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    Quote Originally Posted by holly123 View Post
    find the maximum and minimum values of the function subject to the given contraint.
    f(x,y)= 8x+3y
    subject to (x-1)^2 + (y+2)^2 = 9

    so i took the gradient of each..
    (8, 3)
    (2x-2, 2y+2)

    and to solve for the max and min, we need to use lagrange multipliers. so the gradient of f is equal to the gradient of g times lambda (some constant)

    i have the equations
    8= lambda(2x-2)
    3=lambda(2y+4)
    (x-1)^2 + (y+2)^2 =9

    how do i solve this system of equations? the answer in the back of the book is kind of crazy...
    x= 1 - 24/sqrt 73
    y= -2- 9/sqrt 73
    lambda= -sqrt 73/6
    From the first two

    x - 1 = \frac{8}{2 \lambda},\;\; y + 2 = \frac{3}{2 \lambda}

    then subs. into the third (x-1)^2 +(y+2)^2 = 9. This give you an equation for \lambda which you can solve.
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  3. #3
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    i'm working on the next problem and having the same problem as above. i have the equations:
    -4x= lambda x
    -18y= 2lambda y
    8= lambda z
    (x^2)/4 + (y^2)/9 + (z^2)/16=1

    how do i solve for x and y?
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  4. #4
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    Quote Originally Posted by holly123 View Post
    i'm working on the next problem and having the same problem as above. i have the equations:
    -4x= lambda x
    -18y= 2lambda y
    8= lambda z
    (x^2)/4 + (y^2)/9 + (z^2)/16=1

    how do i solve for x and y?
    From the first you have (\lambda+4) x = 0 and from the second 2(\lambda + 9) y = 0 which gives rise to three cases:

     <br />
(i) \; x = 0, \;y= 0<br />

     <br />
(ii) \; x = 0,\; \lambda = -9,<br />

     <br />
(iii)\; y = 0,\; \lambda = -4.<br />

    Look at each separately.
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    so the first one would be x=o or lambda= -4
    then lambda=-9 or y=0
    lambda= 8/z

    is that right? then what do i do? sorry this part is just not clicking for me...
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  6. #6
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    okay so i tried plugging into the last equation
    so 0 + 0 + (8/lambda)^2/16 = 1
    so lambda= +- 4
    is this correct or totally wrong?
    Last edited by holly123; October 21st 2009 at 05:26 PM.
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