Chapter 4.6, Problem 22E

To determine

To calculate: The dimension of a cylindrical can without a top is made to contain $V{\text{cm}}^3$ of liquid, that will minimize the cost of the metal to make the can.

Answer to Problem 22E

The dimension of a cylindrical can without a top is made to contain $V{\text{cm}}^3$ of liquid is ${\left(\frac{V}{\pi }\right)}^{\frac{1}{3}}$

Explanation of Solution

Given information:

A cylindrical can without a top is made to contain $V{\text{cm}}^3$ of liquid.

Formula used:

Let $r$ be the radiusand $h$ be the height of an open cylindrical can .

Volume of an open cylindrical can is $V=\pi r^{2}h$

Surface area of an open cylindrical can $S=\pi r^2+2\pi rh$

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

A cylindrical can without a top is made to contain $V{\text{cm}}^3$ of liquid

Let $r$ be the radius and $h$ be the height of an open cylindrical can .

Volume of an open cylindrical can is

  $V=\pi r^{2}h$

Rewrite the equation in terms of $h$

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

Let $r$ be the radius and $h$ be the height of an open cylindrical can .

Surface area of an open cylindrical can

  $S=\pi r^2+2\pi rh$

Substitute $h=\frac{V}{\pi r^2}$ ,and simplified

  $\begin{array}{l}S\left(r\right)=\pi r^2+2\pi rh\\ =\pi r^2+2\pi r\times \frac{V}{\pi r^2}\\ S\left(r\right)=\pi r^2+\frac{2V}{r}\end{array}$

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.
3. • 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,

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

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

  $\begin{array}{l}2\pi r-\frac{2V}{r^2}=0\\ \Rightarrow 2\pi r=\frac{2V}{r^2}\\ \Rightarrow r^3=\frac{V}{\pi }\end{array}$

Take cube root on both sides,

  $r={\left(\frac{V}{\pi }\right)}^{\frac{1}{3}}$

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

  $S^{″}\left(r\right)=2\pi +\frac{4V}{r^3}$

Therefore,

  $S^{″}\left(r\right)=2\pi +\frac{4V}{r^3}\ge 0$

For $r={\left(\frac{V}{\pi }\right)}^{\frac{1}{3}}$ the surface area of an open cylindrical can is minimum.

Substitute $r={\left(\frac{V}{\pi }\right)}^{\frac{1}{3}}$ ,to get

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

Conclusion:

The dimension of a cylindrical can without a top is made to contain $V{\text{cm}}^3$ of liquid is ${\left(\frac{V}{\pi }\right)}^{\frac{1}{3}}$