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Thread: Question on surface area of cone

  1. #1
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    Question on surface area of cone

    A manufacturer intends to use the least amount of material to make a hollow right circular cone of radius r cm and vertical height h cm . The cone must be able to contain 100 $\displaystyle cm^3$ of chocolate.

    a) Show that the amount of material used or the curved surface area of the cone is $\displaystyle S=\frac{\sqrt{\pi^2 r^6+90000}}{r}$

    b) Find the radius of the cone so that the amoint of material used will be the minimum. Give your answer in terms of $\displaystyle \pi$

    c) The height of the cone in (b) can be expressed as $\displaystyle h=\sqrt{ar}$, where a is a constant. Find the value of a

    I am able to do (a) and (b), but can't do part (c). Can anyone help??
    Last edited by acc100jt; Aug 16th 2007 at 07:55 PM.
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  2. #2
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    Quote Originally Posted by acc100jt View Post
    A manufacturer intends to use the least amount of material to make a hollow right circular cone of radius r cm and vertical height h cm . The cone must be able to contain 100 $\displaystyle cm^3$ of chocolate.

    a) Show that the amount of material used or the curved surface area of the cone is $\displaystyle S=\frac{\sqrt{\pi^2 r^6+90000}}{r}$

    b) Find the radius of the cone so that the amoint of material used will be the minimum. Give your answer in terms of $\displaystyle \pi$

    c) The height of the cone in (b) can be expressed as $\displaystyle h=\sqrt{ar}$, where a is a constant. Find the value of a

    I am able to do (a) and (b), but can't do part (c). Can anyone help??
    This is good for practice.

    Given: right circular cone, V=100 cu.cm, radius r, height h.
    V = pi(h/3)(r^2) = 100
    h = 300/(pi*r^2) --------(1)

    a) The surface area S.
    S = (length of inclined height)(circumference at mid-height of the cone) <--- using Pappus' Centroid Theorem.
    S = (sqrt[r^2 +h^2])(2pi[r/2])
    S = (sqrt[r^2 +90000/(pi^2 *r^4)])(pi*r)
    S = (sqrt[pi^2 *r^6 +90000] /(pi*r^2))(pi*r)
    S = (sqrt[pi^2 *r^6 +90000])/(r) -----------------shown.

    ----------------------------------------------------------------------
    b) Find r when S is minimum.
    So, the r when dS/dr = 0.

    For simplicity, let's simplify S, let's put the denominator r into the radical.
    S = sqrt[(pi^2 *r^4) +(90000/(r^2))]
    dS/dr = (1/2)[4(pi^2)(r^3) -2(90000)/(r^3)] /sqrt[(pi^2 *r^4) +(90000/(r^2))]
    Set that to zero,
    0 = 2(pi^2)(r^3) -90000/(r^3)
    90000 = 2(pi^2)(r^6)
    r^6 = 45000/(pi^2)
    r = [45000/(pi^2)]^(1/6)
    r = 4.072103 cm.

    In terms of pi,
    r = (4.072103)(pi/pi) = 1.29619 pi -----------answer.

    ------------------------------------------------------------
    c) If h in (b) above is sqrt(a*r), find "a".

    h = 300/(pi*r^2) --------(1)
    h = 300/(pi *[(4.072103)^2)]
    h = 5.758825 cm.
    So,
    5.758825 = sqrt[a(4.072103)]
    Square both sides, and then,
    a = 33.164065 / 4.072103
    a = 8.144211 ------------------------answer.
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  3. #3
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    Thanks a lot.
    The answer given in the exercise book for part (c) is 2, i.e. $\displaystyle a=2$.
    But I couldn't get $\displaystyle a=2$, now you have confirmed my answer is correct. Thanks!!!
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  4. #4
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    Hello, acc100jt!

    This is how I did part (a) . . .


    A manufacturer intends to use the least amount of material to make
    a hollow right circular cone of radius $\displaystyle r$ cm and vertical height $\displaystyle h$ cm .
    The cone must be able to contain 100 cm³ of chocolate.

    a) Show that the amount of material used or the curved surface area
    of the cone is: . $\displaystyle S\:=\:\frac{\sqrt{\pi^2 r^6+90000}}{r}$

    The side view of the cone looks like this:
    Code:
                   r
          * - - + - - *
           \    |    /
            \   |   /
             \  |h / R
              \ | /
               \|/
                *
    $\displaystyle r$ is the radius of the cone, $\displaystyle h$ is the height of the cone.
    The slant height of the cone is: .$\displaystyle R \:=\:\sqrt{r^2+h^2}$ .[1]

    Since the volume is 100 cm³: .$\displaystyle V \:=\:\frac{1}{3}r^2h \:=\:100\quad\Rightarrow\quad h \:=\:\frac{300}{\pi r^2}$

    Substitute into [1]: .$\displaystyle R \:=\:\sqrt{r^2 + \left(\frac{300}{\pi r^2}\right)^2} \:=\:\sqrt{r^2+\frac{90,000}{\pi^2r^4}} :=\:\sqrt{\frac{\pi^2r^6 + 90,000}{\pi^2r^4}}$

    . . Hence: .$\displaystyle R\:=\:\frac{\sqrt{\pi^2r^6 + 90,000}}{\pi r^2} $ .[2]


    The cone is formed from a sector of the circle
    . . with radius $\displaystyle R$ and central angle $\displaystyle \theta$.
    Code:
                  * * *
              *           *
            *               *
           *                 *
    
          *                   *
          *         *         *
          *       * θ *       *
                *       * R
           *  *           *  *
            *               *
              *           *
                  * * * 
                    Rθ
    The area of the sector is: .$\displaystyle S \:=\:\frac{1}{2}R^2\theta$ .[3]

    The arclength of the sector $\displaystyle (R\theta)$,
    . . is the circumference of the base of the cone $\displaystyle (2\pi r)$.
    Hence: .$\displaystyle R\theta \:=\:2\pi r\quad\Rightarrow \quad\theta \:=\:\frac{2\pi r}{R}$

    Substitute into [3]: .$\displaystyle S \:=\:\frac{1}{2}R^2\left(\frac{2\pi r}{R}\right)\quad\Rightarrow\quad S \:=\:\pi rR$ .[4]

    Substitute [2] into [4]: .$\displaystyle S \:=\:\pi r\cdot\frac{\sqrt{\pi^2r^6 + 90,000}}{\pi r^2} $

    Therefore: .$\displaystyle \boxed{S \;=\;\frac{\sqrt{\pi^2r^6 + 90,000}}{r}}$

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