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Maximum volume question

  1. Aug 7, 2007 #1
    Prove that the maximum volume of a rectangular parallelepiped with a fixed surface area is when it is a cube.

    Volume = v, length = l, width = w, height = h and surface area = s. Then we have: v = l*w*h = h^3. s = 2*w*l + 2*w*h + 2*h*l = constant = 6*h^2.

    For a maximum to exist [tex] (\frac{{\partial ^2 y}}{{\partial x^2 }})(\frac{{\partial^2 y}}{{\partial x^2}}) - (\frac{{\partial^2 y}}{{\partial x}{\partial t }}) [/tex] > 0.

    Where x, y and t are unknown to me. I just need some help to get started. Thank you very much.
     
  2. jcsd
  3. Aug 7, 2007 #2

    bel

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    I think you might try using the method of Lagrange multipliers, maximizing [tex] f(l, w, h) = v [/tex] under the condtion that [tex] \phi (l, w, h)= s [/tex] where v and s are constants with the multiplier [tex] \lamda [/tex].
     
    Last edited: Aug 7, 2007
  4. Aug 8, 2007 #3
    The book that I am using is an introduction to Calculus, which does not have Lagrange multipliers in it. So by asking this question, it would imply that the question can be done without recourse to Lagrange multipliers. So I would think.
    Anyway thanks for the reply.
     
  5. Aug 8, 2007 #4

    HallsofIvy

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    Okay, You know that the surface area of a given x by y by z parallelogram is 2xy+ 2xz+ 2yz= A and the volume is V= xyz.
    From 2xy+ 2xz+ 2yz= A, you know that z(2x+2y)= A- 2xy so z= (A- 2xy)/(2x+ 2y). That means that, as a function of x and y, V= xy((A- 2xy)/(2x+ 2y))

    Now, what does that tell you?
     
  6. Aug 10, 2007 #5
    I can do the following: [tex] \frac{{\partial V}}{{\partial x }} = 0 \\[/tex] and [tex] \frac{{ \partial V}}{{\partial y}} = 0 \\[/tex], and then solve for x and y. To find that [tex] x^2 = Y^2 \\[/tex]. Then I can find an expression for V in terms of z and y only, then [tex]\frac{{ \partial V}}{{\partial z}} = 0 \\ [/tex] and solve for y and z to get: [tex] y^2 = z^2 [/tex] Thanks for the reply.
     
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