Chapter 4.6, Problem 19E

To determine

To calculate: The largest possible volume of a cylinder which is inscribed in a sphere of radius $r$

Answer to Problem 19E

The largest possible volume of a cylinderis $\frac{4}{3\sqrt{3}}\pi r^3$ .

Explanation of Solution

Given information:

A right circular cylinder which is inscribed in a sphere of radius $r$

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:

  ${\left(H\right)}^2={\left(P\right)}^2+{\left(B\right)}^2$

Let $h$ be the height and $x$ be the radius of the right circular cylinder.

The volume of the right circular cylinder $V=\pi x^{2}h$

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 the a right circular cylinder which is inscribed in a sphere of radius $r$

  

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:

  ${\left(H\right)}^2={\left(P\right)}^2+{\left(B\right)}^2$

In right angle triangle $△AEO$

  $\begin{array}{l}{r}^2={\left(\frac{h}{2}\right)}^2+x^2\\ \Rightarrow x^2=r^2-{\left(\frac{h}{2}\right)}^2\end{array}$

Recall that, Let $h$ be the height and $x$ be the radius of the right circular cylinder.

The volume of the right circular cylinder

  $V=\pi x^{2}h$

Substitute $x^2=r^2-{\left(\frac{h}{2}\right)}^2$ and simplified

  $\begin{array}{l}V\left(h\right)=\pi \times \left\{r^2-{\left(\frac{h}{2}\right)}^2\right\}\times h\\ =\pi \times \left\{r^2-\frac{h^2}{4}\right\}\times h\\ =\pi \left(h{r}^2-\frac{h^3}{4}\right)\mathrm{......}\left(1\right)\end{array}$

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,

  $V^{\prime }\left(h\right)=\pi \left(r^2-\frac{3h^2}{4}\right)\mathrm{......}\left(2\right)$

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

  $\begin{array}{l}\pi \left(r^2-\frac{3h^2}{4}\right)=0\\ \Rightarrow r^2-\frac{3h^2}{4}=0\\ \Rightarrow r^2=\frac{3h^2}{4}\\ \Rightarrow 3h^2=4r^2\\ \Rightarrow h^2=\frac{4r^2}{3}\end{array}$

Take square root on both sides, to get

  $h=\pm \frac{2r}{\sqrt{3}}$

The equation has two real solutions but length can’t be negative,

Therefore,

  $h=\frac{2r}{\sqrt{3}}$

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

  $\begin{array}{l}{V}^{″}\left(h\right)=-\frac{6h}{4}\\ V^{″}\left(h\right)=-\frac{3h}{2}\end{array}$

Substitute $h=\frac{2r}{\sqrt{3}}$ and simplified

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

Radius can’t be negative,

Therefore,

  $V^{″}\left(h\right)=-\sqrt{3}r<0$

For $h=\frac{2r}{\sqrt{3}}$ volume of right circular cylinder is largest

Now, Substitute $h=\frac{2r}{\sqrt{3}}$ in equation $\left(1\right)$ and simplified

  $\begin{array}{l}V\left(h\right)=\pi \left\{\frac{2r}{\sqrt{3}}\times r^2-\frac{1}{4}\times {\left(\frac{2r}{\sqrt{3}}\right)}^3\right\}\\ =\pi \left\{\frac{2r^3}{\sqrt{3}}-\frac{1}{4}\times \frac{8r^3}{3\sqrt{3}}\right\}\\ =\pi \left\{\frac{2r^3}{\sqrt{3}}-\frac{2r^3}{3\sqrt{3}}\right\}\\ =\pi \left\{\frac{6r^3-2r^3}{3\sqrt{3}}\right\}\\ =\frac{4}{3\sqrt{3}}\pi r^3\end{array}$

Conclusion:

The largest possible volume of a cylinder is $\frac{4}{3\sqrt{3}}\pi r^3$ .