Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Multivariable calculus -critical points

  1. Jul 29, 2006 #1
    A rectangular box is to have a volume of [tex]96m^3[/tex]. The 4 sides and the top are to be made of a material of density [tex]1kg/m^2[/tex] and the flat base is to have a density of [tex]2kg/m^2[/tex]. Suppose that the base is of size x meters by y meters.

    It was found in the first question that the mass, M in kg of the box is:
    [tex]M(x,y) = 3xy + \frac{192}{x} + \frac{192}{y}[/tex]

    Find the dimensions of the box so that the total mass of the box is a minimum.
    So, I just find the critical points, which occur at [tex]f_x = f_y = 0[/tex]
    [tex]f_x = 3y - \frac{192}{x^2} = 0[tex]
    [tex]y = \frac{64}{x^2}[/tex]

    [tex]f_y = 3x - \frac{192}{y^2} = 0[/tex]
    [tex]x(x^3 - 1) = 0,\ \ x = 0,\ 1[/tex]

    So critical points occur at (1, 64) and x = 0. We cannot have a dimension of 0, so minimum mass at x = 1, y = 64. The next question asks to prove that the dimensions found correspond to a minimum value of M(x,y).

    So I use second derivatives and the Hessian matrix.
    [tex]f_{xx} = \frac{384}{x^3},\ \ f_{xy} = 3,\ \ f_{yy} = \frac{384}{y^3}[/tex]
    [tex]\mbox{det(H)} = \frac{384^2}{(xy)^3} - 9[/tex]
    At (1, 64), det(H) < 0 which means that it is a saddle point, not a minimum. So, where have I gone wrong, or is the answer simply "there is no minimum"?
     
    Last edited: Jul 29, 2006
  2. jcsd
  3. Jul 29, 2006 #2

    benorin

    User Avatar
    Homework Helper

    a little help, too much perhaps

    Your critical points aren't right: notice how m(x,y)=m(y,x) so that any critical points will have x=y? well, anyhow up to here it is ok:

    [tex]y=\frac{64}{x^2}, x=\frac{64}{y^2}[/tex]​

    now assume [tex]x\neq 0,y\neq 0[/tex] (because of just what you said: can't have 0 volume), now simplify them to get

    [tex]x^2y=64, xy^2=64[/tex]​

    multiply the first equation by y and the second by x to get

    [tex]x^2y^2=64y, x^2y^2=64x[/tex]​

    since both equations have one side equal to [tex]x^2y^2[/tex] we can set them equal to eachother as in [tex]64y=64x[/tex] so x=y, plug this into any previous equation, say [tex]x^2y=64[/tex] to get [tex]x^3=64[/tex] so x=4, but x=y hence x=y=4 is the critical point.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook