Chapter 4.6, Problem 27E

To determine

To calculate: The maximum capacity of a cup which is made from a circular piece of paper of radius $R$ by cutting out a sector and joining the edge $CA\text{and }CB$

Answer to Problem 27E

The maximum capacity of a cup which is made from a circular piece of paper of radius $R$ by cutting out a sector and joining the edge $CA\text{and }CB$ is $\frac{2\pi R^3}{9\sqrt{3}}$

Explanation of Solution

Given information:

A cup which is made from a circular piece of paper of radius $R$ by cutting out a sector and joining the edge $CA\text{and }CB$

.

Formula used:

Pythagorean Theorem: The sum of the squares on the legs of the right angled triangle is equal to the square on the side opposite to the right angle triangle. That is:

  

Let $r$ be the radius and $h$ be the height of the cone. Then

The volume of the cone is $V=\frac{1}{3}\pi r^{2}h$

And

Let $f$ be a differentiable function defined on an interval $I$ and let $a\in I$ .

Then

1. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from positive to negative as $x$ increases through $a$ , i.e. if $f^{\prime }(x)>0$ at every point sufficiently close to and to the left of $a$ , and $f^{\prime }(x)<0$ at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local maxima
2. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from negative to positive as $x$ increases through $a$ , i.e. if $f^{\prime }(x)<0$ at every point sufficiently close to and to the left of $a$ , and at $f^{\prime }(x)>0$ every point sufficiently close to and to the right of $a$ , then $a$ is a point of local minima.
3. $f^{\prime }(a)=0$ and If $f^{\prime }(x)$ does not change sign as $x$ increases through $a$ , then $a$ is neither a point of local maxima nor a point of local minima.
4. • If $f^{″}(a)>0$ then $f$ has a local minimum at $x=a$
   • If $f^{″}(a)<0$ then $f$ has a local maximum at $x=a$

Calculation:

As per the given problem

Draw the diagram of a cup which is made from a circular piece of paper of radius $R$ by cutting out a sector and joining the edge $CA\text{and }CB$

  

Recall that, Pythagorean Theorem: The sum of the squares on the legs of the right angled triangle is equal to the square on the side opposite to the right angle triangle.

In $△ADC$

  $\begin{array}{l}{R}^2=r^2+h^2\\ \Rightarrow r^2=R^2-h^2\end{array}$

Recall that, Let $r$ be the radius and $h$ be the height of the cone. Then

The volume of the cone is $V=\frac{1}{3}\pi r^{2}h$

Substitute $r^2=R^2-h^2$ ,and simplified

  $\begin{array}{l}V=\frac{1}{3}\pi r^{2}h\\ =\frac{1}{3}\pi \left(R^2-h^2\right)h\\ =\frac{1}{3}\pi \left(R^{2}h-h^3\right)\mathrm{......}\left(1\right)\end{array}$

Recall that,

Let $f$ be a differentiable function defined on an interval $I$ and let $a\in I$ .

Then

1. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from positive to negative as $x$ increases through $a$ , i.e. if $f^{\prime }(x)>0$ at every point sufficiently close to and to the left of $a$ , and $f^{\prime }(x)<0$ at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local maxima
2. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from negative to positive as $x$ increases through $a$ , i.e. if $f^{\prime }(x)<0$ at every point sufficiently close to and to the left of $a$ , and at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local minima.
3. $f^{\prime }(a)=0$ and If $f^{\prime }(x)$ does not change sign as $x$ increases through $a$ , then $a$ is neither a point of local maxima nor a point of local minima.
4. • If $f^{″}(a)>0$ then $f$ has a local minimum at $x=a$
   • If $f^{″}(a)<0$ then $f$ has a local maximum at $x=a$

Differentiate on both sides,

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

Solve for $V^{\prime }\left(h\right)=0$ , and simplified

  $\begin{array}{l}\frac{1}{3}\pi \left(R^2-3h^2\right)=0\\ \Rightarrow R^2-3h^2=0\\ \Rightarrow h^2=\frac{R^2}{3}\end{array}$

Take square root on both sides,

  $h=\pm \frac{R}{\sqrt{3}}$

But length never negative,

Therefore,

  $h=\frac{R}{\sqrt{3}}$

Differentiate equation $\left(2\right)$ with respect to $l$

  $\begin{array}{l}{V}^{″}\left(h\right)=\frac{1}{3}\pi \times -6h\\ =-2\pi h\end{array}$

Therefore,

  $V^{″}\left(h\right)=-2\pi h<0$

For $h=\frac{R}{\sqrt{3}}$ the volume of cone is maximum

Substitute $h=\frac{R}{\sqrt{3}}$ in equation $\left(1\right)$ and simplified

  $\begin{array}{l}V\left(\frac{R}{\sqrt{3}}\right)=\frac{1}{3}\pi \left\{R^2\times \frac{R}{\sqrt{3}}-{\left(\frac{R}{\sqrt{3}}\right)}^3\right\}\\ =\frac{1}{3}\pi \left\{\frac{R^3}{\sqrt{3}}-\frac{R^3}{3\sqrt{3}}\right\}\end{array}$

Take least common multiple of $\sqrt{3}\text{and 3}\sqrt{3}$ is $\text{3}\sqrt{3}$

  $\begin{array}{l}=\frac{1}{3}\pi \left\{\frac{3R^3-R^3}{3\sqrt{3}}\right\}\\ =\frac{1}{3}\pi \times \frac{2R^3}{3\sqrt{3}}\\ =\frac{1}{3}\pi \times \frac{2R^3}{3\sqrt{3}}\\ =\frac{2\pi R^3}{9\sqrt{3}}\end{array}$

Conclusion:

Thus the maximum capacity of a cup which is made from a circular piece of paper of radius $R$ by cutting out a sector and joining the edge $CA\text{and }CB$ is $\frac{2\pi R^3}{9\sqrt{3}}$