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Find the minimum volume of a cylinder of revolution given constraints

by RoRi

Find the largest cylinder (in terms of volume) we can obtain by revolving a rectangle lying on the x-axis and contained entirely in the region of the plane bounded by the axes and the curve

    \[ y = \frac{x}{x^2 + 1}. \]

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First, we know that the two upper corners must have the same distance from the x-axis (since this is a rectangle the line joining the upper corners must be horizontal). Both upper corners will lie on the curve y since if either corner were not on the curve we could obtain a larger rectangle (hence a larger cylinder when we rotate) by extending the rectangle so the corner is on the curve. Let a be the the x coordinate of the upper left corner, and b be the x coordinate of the upper right corner. Since these both lie on the curve we have,

    \begin{align*} \frac{a}{1+a^2} = \frac{b}{1+b^2} && \implies && a(1+b^2) &= b(1+a^2) \\ && \implies && a + a b^2 - b - a^2 b &= 0 \\ && \implies && a^2(-b) + a(1 + b^2) - b &= 0 \\ && \implies && a &= \frac{-(1+b^2) \pm \sqrt{(1+b^2)^2 - 4b^2}}{-2b} \\ && \implies && a &= \frac{1+b^2 \pm \sqrt{1 - 2b^2 + b^4}}{2b} \\ && \implies && a &= \frac{1 + b^2 \pm (1-b^2)}{2b} \end{align*}

However, one of these solutions, \frac{1+b^2 - (1-b^2)}{2b} is the trivial case that a = b, so the cylinder in this case has volume 0. The other solution then is

    \[ a = \frac{1 + b^2 + 1 - b^2}{2b} = \frac{1}{b}. \]

So, then we compute the volume of the cylinder,

    \begin{align*} V = \pi r^2 h &= \pi (f(x))^2 \left( x - \frac{1}{x}\right) \\ &= \pi \left( \frac{x^2}{(1+x^2)^2} \right) \left( \frac{x^2-1}{x} \right) \\ &= \pi \frac{x^3 - x}{(x^2+1)^2}. \end{align*}

Taking the derivative,

    \begin{align*} \frac{dV}{dx} &= \pi \frac{(3x^2-1)(x^2+1)^2 - (x^3 -x )(4x)(x^2+1)}{(x^2+1)^4} \\ &= \pi \frac{-x^4 + 6x^2 - 1}{(x^2+1)^3}. \end{align*}

Setting this derivative equal to 0, we then have

    \begin{align*} \pi \frac{-x^4 + 6x^2 - 1}{(x^2+1)^3} = 0 && \implies && x^4 - 6x^2 + 1 &= 0 \\ && \implies && x^2 &= 3 \pm 2 \sqrt{2}. \end{align*}

These two values of x^2 then correspond to the values for a and b, so we have

    \[ b = \sqrt{x} = \sqrt{3 + 2\sqrt{2}} = 1 + \sqrt{2}. \]

(Where we can eliminate the negative value of x since the problem requires the value to be positive.) Therefore, the maximum volume V_{\text{max}} is given by

    \begin{align*} V_{\text{max}} &= \pi \frac{(1+\sqrt{2})^3 - 1 - \sqrt{2}}{(4 + \sqrt{2})^2} \\ &= \pi \left( \frac{6 + 4 \sqrt{2}}{24 + 16\sqrt{2}} \right) \\ &= \pi \left( \frac{3 + 2 \sqrt{2}}{12 + 8 \sqrt{2}} \right) \\ &= \pi \left( \frac{ (3+2 \sqrt{2} )(12 - 8 \sqrt{2})}{16}\right) \\ &= \pi \left( \frac{4}{16} \right) \\ &= \frac{\pi}{4}. \end{align*}

One comment

  1. Jan says:

    – Both values of x are pos.
    – Your x value gives minimum volume, i.e. zero.
    – Maybe your answer is correct because the y is the same on both ends of the rectangle.

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