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Thread: [SOLVED] Applications of Differentiation Question

  1. #1
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    [SOLVED] Applications of Differentiation Question

    Question:
    A hollow circular cylinder, open at one end, is constructed of thin sheet metal. The total external surface area of the cylinder is $\displaystyle 192\pi$$\displaystyle cm^2$. The cylinder has a radius of $\displaystyle r$$\displaystyle cm$ and a height of $\displaystyle h$$\displaystyle cm$.

    (i) Express h in terms of r and show that the volume, $\displaystyle V$$\displaystyle cm^3$, of the cylinder is given by:
    $\displaystyle V = \frac{1}{2}\pi(192r - r^3)$

    Given that $\displaystyle r$ can vary,
    (ii) Find the value of $\displaystyle r$ for which $\displaystyle V$ has a stationary value,
    (iii) find this stationary value and determine whether it is a maximum or a minimum.


    (i) Can someone please show me how to get $\displaystyle V = \frac{1}{2}\pi(192r - r^3)$?
    Thanks from SilVarx
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    Member Danshader's Avatar
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    external surface area of a closed cylinder
    = $\displaystyle 2\pi r h + 2\pi r^2$

    external surface area for one open end of cylinder
    = $\displaystyle 2\pi r h + \pi r^2$
    = $\displaystyle 192\pi$

    obtain an expression for h.

    volume of cylinder
    = $\displaystyle \pi r^2 h$

    substitute the value of h into the volume equation
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  3. #3
    Member looi76's Avatar
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    Quote Originally Posted by Danshader View Post
    external surface area of a closed cylinder
    = $\displaystyle 2\pi r h + 2\pi r^2$

    external surface area for one open end of cylinder
    = $\displaystyle 2\pi r h + \pi r^2$
    = $\displaystyle 192\pi$

    obtain an expression for h.

    volume of cylinder
    = $\displaystyle \pi r^2 h$

    substitute the value of h into the volume equation
    Thanks Danshader

    $\displaystyle 2\pi{rh} + \pi{r^2} = 192\pi$

    $\displaystyle \frac{2\pi{rh}}{\pi} + \frac{\pi{r^2}}{\pi} = \frac{192\pi}{\pi}$

    $\displaystyle 2rh + r^2 = 192$

    $\displaystyle 2rh = 192 - r^2$

    $\displaystyle h = \frac{192 - r^3}{2r}$

    $\displaystyle = \pi{r^2h}$

    $\displaystyle =\pi{r^2}\frac{192 - r^3}{2r}$

    $\displaystyle V = \frac{1}{2}\pi{}(192r - r^3)$

    Now how do I find the value of r?
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  4. #4
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    Definition:A point 'u' at which the derivative of a function f(x) vanishes,i.e. f'(u) = 0 is called a stationary point.


    This means you have to find values of r such that V'(r) = 0.
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    Member Danshader's Avatar
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    yup just like what isomorphism said, a stationary point can be found when the derivative of V with respect to r is 0

    $\displaystyle \frac{dV}{dr} = 0$

    earlier you already prove the equation of the volume:

    $\displaystyle V = \frac{1}{2}\pi{}(192r - r^3)$


    hence all you need to do is to differentiate this equation with respect to r and equate it to 0:

    $\displaystyle \frac{d}{dr}{(\frac{1}{2}\pi{}(192r - r^3)}) = 0$

    solve for r. good luck
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