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Q29E

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Found in: Page 239

### Essential Calculus: Early Transcendentals

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, $$V = \frac{{2\pi {R^3}}}{{9\sqrt 3 }}$$.

See the step by step solution

## Step 1: To sketch the figure and to mention given data

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}

## Step 2: To find the height of the cone

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}

## Step 3: To find the maximum capacity of the cup

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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