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Q29E

Expert-verifiedFound in: Page 239

Book edition
2nd

Author(s)
James Stewart

Pages
830 pages

ISBN
9781133112280

**A cone-shaped drinking cup is made from a circular piece of paper of radius \(R\) by cutting out a sector and joining the edges \(CA\) and \(CB\). Find the maximum capacity of such a cup.**

The maximum capacity of the cup is, \(V = \frac{{2\pi {R^3}}}{{9\sqrt 3 }}\).

Let \(r\) be the radius of the cone and \(h\) be the height of the cone.

Let \(R\) be the radius of the original circular piece of paper.

The sketch is shown in the figure below:

Using Pythagorean theorem, we have,

\(\begin{aligned}{l}{r^2} + {h^2} &= {R^2}\\ \Rightarrow {r^2} &= {R^2} - {h^2}\end{aligned}\)

Volume of cone is given by,

\(V = \frac{1}{3}\pi {r^2}h\,\,\, \cdots \cdots \left( 1 \right)\)

Using the above value in \(\left( 1 \right)\) , we get,

\(\begin{aligned}{l}V &= \frac{1}{3}\pi \left( {{R^2} - {h^2}} \right)h\\ \Rightarrow V &= \frac{1}{3}\pi \left( {{R^2}h - {h^3}} \right)\,\,\, \cdots \cdots \left( 2 \right)\end{aligned}\)

Differentiating \(\left( 2 \right)\) , we get,

\(V' = \frac{1}{3}\pi \left( {{R^2} - 3{h^2}} \right)\).

Thus, solving for \(V' = 0\), we get,

\(\begin{aligned}{c}\frac{1}{3}\pi \left( {{R^2} - 3{h^2}} \right) &= 0\\{R^2} - 3{h^2} &= 0\\{R^2} &= 3{h^2}\\{h^2} &= \frac{{{R^2}}}{3}\\h &= \frac{R}{{\sqrt 3 }}\end{aligned}\)

Putting this value in \(\left( 2 \right)\), we get,

\(\begin{aligned}{c}V &= \frac{1}{3}\pi \left( {{R^2}\frac{R}{{\sqrt 3 }} - {{\left( {\frac{R}{{\sqrt 3 }}} \right)}^3}} \right)\,\\V &= \frac{{\pi {R^3}}}{{3\sqrt 3 }} - \frac{{\pi {R^3}}}{{9\sqrt 3 }}\\V &= \frac{{2\pi {R^3}}}{{9\sqrt 3 }}\end{aligned}\)

Hence, the maximum capacity of the cup is, \(V = \frac{{2\pi {R^3}}}{{9\sqrt 3 }}\).

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