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Math Help - Applications of maxima / minima: inscribed Cylinder

  1. #1
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    Applications of maxima / minima: inscribed Cylinder

    a.)Estimate the dimensions of a right circular cylinder of greatest lateral surface area that can be incribed in a sphere with a radius of 6in. Confirm your estimates analytically.

    b.)Estimate the dimensions of a right circular cylinder of Greatest Volume that can be inscribed in a sphere w/ radius 6in. Confirm your estimates.

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  2. #2
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    Hello, ^_^Engineer_Adam^_^!

    I'll start this one . . .


    a) Find the dimensions of a right circular cylinder of greatest lateral
    surface area that can be inscribed in a sphere with a radius of 6 in.

    The lateral surface area of a cylinder is: .S .= .2πrh
    . . where r is the radius and h is the height.
    Code:
                  * * *
              *     :     *
            *-------+-------*
           *|       :       |*
            |       :       |
          * |       :       | *
          * |       *       | *
          * |       | \ 6   | *
            |      y|   \   |
           *|       |  x  \ |*
            *-------+------ *
              *     |     *
                  * * *

    We see that the radius is x and the height is 2y.

    We also see that: .x + y .= .36 . . y .= .(36 - x)^


    And we have: .S .= .2πrh .= .2πx(2y) .= .4πx(36 - x)^

    Can you finish it now?

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  3. #3
    MHF Contributor
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    a.)Estimate the dimensions of a right circular cylinder of greatest lateral surface area that can be incribed in a sphere with a radius of 6in. Confirm your estimates analytically.

    b.)Estimate the dimensions of a right circular cylinder of Greatest Volume that can be inscribed in a sphere w/ radius 6in. Confirm your estimates.

    Lateral Surface of right cylinder = 2pi(radius)(height)
    Volume of right cylinder = pi(radius^2)(height)

    Draw the figure.
    Draw a circle. Draw the inscribed right circular cylinder vertically. Draw the vertical diameter of the circle. Draw a radius to any of the four corners of the "rectangle". A right triangle is formed, with these:
    ---hypotenuse = 6 in.
    ---vertical leg = (1/2)height of cylinder = h/2
    ---horizontal leg = radius of cylinder = r
    ---central angle, between hypotenuse and vertical leg = angle X

    r = 6sinX

    h/2 = 6cosX
    h = 12cosX

    -----------------------------------
    1) Greatest lateral surface, A.

    A = 2pi(r)h
    A = 2pi(6sinX)(12cosX)
    A = 36pi(2sinXcosX)
    A = 36pi(sin(2X))
    dA/dX = 36pi[cos(2X) *2]
    dA/dX = 72pi[cos(2X)]
    Set that to zero,
    0 = 72pi[cos(2X)]
    0 = cos(2X)
    2X = arccos(0) = pi/2
    X = pi/4 ----------------------for max A

    Then,
    r = 6sin(pi/4) = 6(sqrt(2) /2) = 3sqrt(2) = 4.24264 in.
    h = 12cos(pi/4) = 12(sqrt(2) /2) = 6sqrt(2) = 8.48528 in.

    Therefore, the inscribed right cylinder of greatest lateral area is 8.48528 inches in diameter and is 8.48528 inches also high. -----------answer.

    -------------------------------------------------
    b) Greatest Volume, V.

    V = pi(r^2)h
    V = pi[(6sinX)^2 ](12cosX)
    V = pi[36sin^2(X)](12cosX)
    V = 12*36pi[(1 -cos^2(X)]cosX
    V = 12*36pi[cosX -cos^3(X)] -------------------(1)
    dV/dX = 12*36pi[-sinX -3cos^2(X)*(-sinX)]
    dV/dX = 12*36pi[(-sinX)[1 -3cos^2(X)]
    Set that to zero,
    (-sinX)[1 -3cos^2(X)] = 0

    -sinX = 0
    0 = sinX
    X = arcsin(0) = 0 or pi -----------***

    1 -3cos^2(X) = 0
    cos^2(X) = 1/3
    cosX = +,-sqrt(1/3) = +,-0.57735
    X = arccos(+,-0.57735) = 0.9553 radians or 2.1863 radians. ----***

    So there are 4 possible values of the angle X for max V. Plug them into (1). Only one of them gives a positive V---the X = 0.9553 radians only.

    So,
    r = 6sin(0.9553) = 4.8989 in. -----diameter = 2r = 9.7978 in.
    h = 12cos(0.9553) = 6.9284 in

    Therefore, the inscribed right cylinder of greatest volume is 9.7978 inches in diameter and is 6.9284 inches high. -----------answer.
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