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Math Help - Applications involving an absolute extremum on a closed interval

  1. #1
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    Applications involving an absolute extremum on a closed interval

    Confirm analytically your estimates using your graphics calculator

    A tent in the shape of a pyramid with a square base is to be constructed from a square piece of material of side 5m. In the base of the pyramid, let x meters be the distance from the center to a side. See the figure.

    a.) On your graphics calculator, find accurate to the nearest one hundredth of a meter the value of x for which the volume is of a pyramid is a maximum.

     V = \frac{1}{3}Bh

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  2. #2
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    Confirm analytically your estimates using your graphics calculator

    A tent in the shape of a pyramid with a square base is to be constructed from a square piece of material of side 5m. In the base of the pyramid, let x meters be the distance from the center to a side. See the figure.

    a.) On your graphics calculator, find accurate to the nearest one hundredth of a meter the value of x for which the volume is of a pyramid is a maximum.

     V = \frac{1}{3}Bh

    Sorry, but I do not know how to use any graphing calculator ---no kidding---so can I solve this analytically instead?

    V = (1/3)Bh
    means Volume = (1/3)(area of square base)(vertical height of the apex)

    We know B = (2x)^2 = 4x^2
    But we don't know h.

    If we put up the pyramidal tent, the h is the vertical component of the slant height of the apex.
    Let y = that slant height.

    On the posted drawing, or when the tent is still flat on the ground or not yet cut, if we draw a horizontal, or vertical, line through the center of the 5m-square material, we see that
    y +2x +y = 5
    So,
    2y = 5 -2x
    y = (2.5 -x) m. ----------***

    Put up the tent again.
    We see an isosceles triangle silhouette. The base is 2x m long. The two equal sides are y meters each. Draw the h.
    By Pythagorean Theorem,
    y^2 = x^2 +h^2
    h = sqrt(y^2 -x^2)
    So,
    h = sqrt[(2.5 -x)^2 -x^2]
    h = sqrt[6.25 -5x +x^2 -x^2]
    h = sqrt(6.25 -5x) ---------------***

    Hence,
    V = (1/3)(4x^2)sqrt(6.25 -5x)
    V = (4/3)(x^2)sqrt(6.25 -5x)
    Differentiate both sides with respect to x,
    dV/dx = (4/3){(x^2)[(1/2)(-5 /sqrt(6.25 -5x)] +sqrt(6.25 -5x)[2x]}
    dV/dx = (-10/3)(x^2)/sqrt(6.25 -5x) +(8/3)(x)sqrt(6.25 -5x)
    Set dV/dx to zero,
    0 = (-10/3)(x^2)/sqrt(6.25 -5x) +(8/3)(x)sqrt(6.25 -5x)
    Multiply both sides by 3sqrt(6.25 -5x),
    0 = -10x^2 +8(6.25 -5x)
    0 = -5x^2 +4(6.25 -5x)
    0 = -5x^2 +25 -20x
    0 = -x^2 -4x +5
    x^2 +4x -5 = 0
    (x+5)(x-1) = 0
    x = -5m or 1m
    There are no negative dimensionsn so reject x = -5m
    So, x = 1 m.

    Therefore for maximum volume of the tent, x = 1 meter. -------answer.

    -----------------------
    Ooppss, mistake. Let me edit it

    ....
    0 = (-10/3)(x^2)/sqrt(6.25 -5x) +(8/3)(x)sqrt(6.25 -5x)
    Multiply both sides by 3sqrt(6.25 -5x),
    0 = -10x^2 +8x(6.25 -5x)
    Divide both sides by 2x,
    0 = -5x +4(6.25 -5x)
    0 = -5x +25 -20x
    0 = -25x +25 -------------------------:-), yeah, really?
    x = 1 m ---------------hey!

    Therefore for maximum volume of the tent, x = 1 meter. -------answer.
    Last edited by ticbol; February 9th 2007 at 09:47 PM.
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  3. #3
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    Good work!
    No wonder... youre a civil engineer!
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