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Math Help - Help Related rates

  1. #1
    Senior Member
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    Help Related rates

    Hi people this is my last problem so sorry for asking too much because its our finals day tommorow...

    1.) A conical tent with no floor is to have a capacity of 1000cubic meters. Find the dimensions that minimize the amount of canvas required:

    2.) A box with a square base has an open top. The area of the material in this box is 100cm^2. What should be the dimensions in order to make the volume as large as possible?

    Just give me formula Ill be differentiating this:

    THanks SOOO much for all the big help!
    My Next bunch of questions will be about integrals (i hope if i pass)
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    Hi people this is my last problem so sorry for asking too much because its our finals day tommorow...

    1.) A conical tent with no floor is to have a capacity of 1000cubic meters. Find the dimensions that minimize the amount of canvas required:
    The curved surface area of a cone is slant height * perimiter of base/2,

    so if the height is h, and the radius of the base is r, we have slant height:

    s = sqrt(h^2+r^2)

    So the area of canvas required is:

    A = sqrt(h^2+r^2) (2 pi r)/2

    The volume is:

    V = (1/3) (pi r^2) h = 1000,

    so:

    h = 3000/(pi r^2),

    and area of canvas:

    A = sqrt([3000/(pi r^2)]^2+r^2) (2 pi r)/2

    But if A is minimise then so is A^2, so you may as well minimise:

    A^2 = ([3000/(pi r^2)]^2+r^2) pi^2 r^2

    RonL
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    2.) A box with a square base has an open top. The area of the material in this box is 100cm^2. What should be the dimensions in order to make the volume as large as possible?
    Area of material is:

    A= b^2 + 4*b*h = 100,

    where h is the height of the box and b the side of the base in cm.

    So:

    h = [100 - b^2]/(4 b) ... (1)

    The volume is:

    V = b^2 h = b [100 - b^2]/4 ... (2)

    So to find the b that maximises the volume we solve dV/db=0 (determining
    which solutions correspond to maxima/minima if necessary), then solve for
    h using (1)

    RonL
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