Chapter 4.6, Problem 28E

To determine

To calculate: The height and radius of a cup that will use the smallest amount of paper. If cone-shaped paper drinking cup is to be made to hold $27{\text{cm}}^3$ of water.

Answer to Problem 28E

The height and radius of a cup that will use the smallest amount of paper are $3.72\text{cm}$ and $2.63\text{cm}$ .

Explanation of Solution

Given information:

A cone-shaped paper drinking cup is to be made to hold $27{\text{cm}}^3$ of water.

.

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 and $l$ be the slant height of the cone. Then

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

Surface area of the cone is $S=\pi rl$

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 cone-shaped paper drinking cup is to be made to hold $27{\text{cm}}^3$ of water.

  

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 $△ADC$

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

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 $V=27$ , and solve for $r^2$ ,

  $\begin{array}{l}27=\frac{1}{3}\pi r^{2}h\\ \Rightarrow r^2=\frac{81}{\pi h}\mathrm{......}\left(1\right)\end{array}$

Recall that, let $r$ be the radius and $h$ be the height and $l$ be the slant height of the cone. Then

Surface area of the cone is

  $S=\pi rl$

Substitute $l=\sqrt{r^2+h^2}$ , to get

  $S=\pi r\sqrt{r^2+h^2}$

Square on both sides,

  $S^2={\pi }^2{r}^2\left(r^2+h^2\right)$

Substitute $r^2=\frac{81}{\pi h}$ , to get

  $\begin{array}{l}{S}^2\left(h\right)={\pi }^2{r}^2\left(r^2+h^2\right)\\ ={\pi }^2\times \frac{81}{\pi h}\left(\frac{81}{\pi h}+h^2\right)\\ =81\pi \left(\frac{81}{\pi h^2}+h\right)\end{array}$

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,

  $\begin{array}{l}\frac{d}{dx}{S}^2\left(h\right)=\frac{d}{dx}\left\{81\pi \left(\frac{81}{\pi h^2}+h\right)\right\}\\ =81\pi \left(-\frac{162}{\pi h^3}+1\right)\mathrm{......}\left(2\right)\end{array}$

Solve for $\frac{d}{dx}{S}^2\left(h\right)=0$ , and simplified

  $\begin{array}{l}81\pi \left(-\frac{162}{\pi h^3}+1\right)=0\\ \Rightarrow \frac{162}{\pi h^3}=1\\ \Rightarrow h^3=\frac{162}{\pi }\end{array}$

Take cube root on both sides,

  $h\approx 3.72$

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

  $\begin{array}{l}\frac{d^2}{d{h}^2}{S}^2\left(h\right)=\frac{d}{dh}\left\{81\pi \left(-\frac{162}{\pi h^3}+1\right)\right\}\\ =81\pi \times \frac{486}{\pi h^4}\\ =\frac{39366}{h^4}\end{array}$

Therefore,

  $\frac{d^2}{d{h}^2}{S}^2\left(h\right)=\frac{39366}{h^4}\ge 0$

For $h\approx 3.72$ the surface area of cone is smallest

Substitute $h\approx 3.72$ in equation $\left(1\right)$ and

  $r^2\approx \frac{81}{\pi \times 3.72}$

Take square root on both sides,

  $r\approx 2.63$

Conclusion:

Thus the height and radius of a cup that will use the smallest amount of paper are $3.72\text{cm}$ and $2.63\text{cm}$ .