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LaGrange max/min (calc 3)

  1. Jun 27, 2008 #1
    1. The problem statement, all variables and given/known data
    i needed to find the max and min possible volume for a box with edges that = 200cm and surface area that = 1500cm^2 using Lagrange multipliers.

    2. Relevant equations
    edges: 4x + 4y + 4z = 200cm
    Area: 2xy + 2xz + 2yz = 1500 cm^2
    Volume = xyz

    3. The attempt at a solution

    i brought it down to this:
    Vol = f
    Area = G
    Edges = g

    ^f = ^G# + ^g$
    (# and & just means 'some number')
    ( ^ is the gradient of the function)
    when i simplify everything out i get these statements:

    yz = #(y + z) + $
    xz = #(x + z) + $
    xy = #(x + y) + $
    x + y +z = 50
    xy + yz + yz = 750

    when i try to replace values it seems to get messy. do i just throw in random values for # and $ or something? im 99% sure that they cannot = 0. that is pretty much all i can say from these statements right now lol
  2. jcsd
  3. Jun 27, 2008 #2


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    Science Advisor
    Homework Helper

    Let's call L=# and M=$. Subtract your first two equation from each other. This gives (x-y)*z=L*(x-y). Or (z-L)*(x-y)=0. This tells you either z=L or x=y. Similarly y=L or x=z, or x=L or y=z. If you think about it, in any case, two of the lengths must be equal. Put this information into your last two equations. Now you have two equations in two unknowns. Not hard to solve at all.
  4. Jun 28, 2008 #3
    ok cool, thanks a lot, found the answer
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