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Essential Calculus: Early Transcendentals
Found in: Page 239
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

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Short Answer

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 by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

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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