Question

A cone-shaped paper drinking cup is to be made to hold \(27{\rm{ c}}{{\rm{m}}^3}\) of water. Find the height and radius of the cup that will use the smallest amount of paper.

Short answer

The required values are:

\(\begin{aligned}{l}h \approx 3.722{\rm{ units}}\\r \approx 2.632{\rm{ units}}\end{aligned}\).

Step-by-step solution

Optimization of a Function

The Optimization of a Function \(f\left( x \right)\) for any real value of \(c\)can be given as:

\(\begin{aligned}{l}f\left( x \right)\left| {_{x = c} \to \max \to \left\{ \begin{aligned}{l}{\rm{if }}f'\left( x \right) > 0\,\,\,\,\,\,\forall x < c\\{\rm{if }}f'\left( x \right) < 0\,\,\,\,\,\,\forall x > c\end{aligned} \right.} \right.\\{\rm{and}}\\f\left( x \right)\left| {_{x = c} \to \min \to \left\{ \begin{aligned}{l}{\rm{if }}f'\left( x \right) < 0\,\,\,\,\,\,\forall x < c\\{\rm{if }}f'\left( x \right) > 0\,\,\,\,\,\,\forall x > c\end{aligned} \right.} \right.\end{aligned}\).

Establishing function for volume and surface area.

The volume of the cone is given as: \(27{\rm{ c}}{{\rm{m}}^3}\).

Now, for any cone of radius\(r\)and height\(h\), the volume is given by:

\(\begin{aligned}{c}V = \frac{\pi }{3}{r^2}h\\\frac{\pi }{3}{r^2}h = 27\\{r^2} = \frac{{81}}{{\pi h}}\end{aligned}\)

Also, surface area is given by:

\(\begin{aligned}{c}S = \pi r\sqrt {{r^2} + {h^2}} \\A = {S^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.............\left\{ {{\rm{let}}} \right\}\\A = {\pi ^2}{r^2}\left( {{r^2} + {h^2}} \right)\\ = {\pi ^2}\frac{{81}}{{\pi h}}\left( {\frac{{81}}{{\pi h}} + {h^2}} \right)\\ = {\left( {\frac{{81}}{h}} \right)^2} + 81\pi h\end{aligned}\)

Optimizing the function according to given condition.

For minimization, we have:

\(\begin{aligned}{c}A'\left( h \right) = 0\\\frac{d}{{dh}}\left( {{{\left( {\frac{{81}}{h}} \right)}^2} + 81\pi h} \right) = 0\\{\left( {\frac{{81}}{h}} \right)^2}\left( { - \frac{2}{h}} \right) + 81\pi = 0\\{h^3} = \frac{{162}}{\pi }\\h \approx 3.722{\rm{ cm}}\end{aligned}\)

Now, we have:

\(A''\left( h \right) > 0\)

Therefore, the function will have absolute minimumvalue at \(h = 3.722{\rm{ cm}}\). For this, we have the radius as:

\(\begin{aligned}{l}{r^2} = \frac{{81}}{{\pi \left( {3.722} \right)}}\\r \approx 2.632{\rm{ cm}}\end{aligned}\)

Hence, these are the required values.