# Design a closed cylindrical container that holds 100 cm^3 and has the minimal possible surface...

## Question:

Design a closed cylindrical container that holds {eq}\rm 100 \ cm^3 {/eq} and has the minimal possible surface area. What should its dimensions be?

## Applications of derivatives to maxima and minima problems:

Derivatives find applications in almost all mathematical formulations of real-world problems. Be it economics, engineering, and other streams.

In maxima and minima problems, derivatives are used to optimize, that is, find the maximum and minimum value of one or more variable dependent on other independent variables.

Let us consider a cylinder of radius r and height h.

It's Volume V, and total surface area A are given in terms of its radius and height as follows;

{eq}\begin{align*} V &= \pi r^2 h \\ A &= 2 \pi r h + 2 \pi r^2. \end{align*} {/eq}

The volume V is given to us.

{eq}\begin{align*} V = 100 \ cm^3 \\ \end{align*} {/eq}

Area A can be written in terms of only one variable using the given value of V.

{eq}\begin{align*} A &= 2 \pi r h + 2 \pi r^2 \\ A &= 2 \pi r \frac { V } { \pi r^2 } + 2 \pi r^2 \\ A &= \frac { 2V } { r } + 2 \pi r^2 \dots(1) \end{align*} {/eq}

Here V is a constant since its value is given to us.

Differentiating both sides of the equation 1 with respect to r ;

{eq}\begin{align*} \frac { d } { dr } A &= \frac { d } { dr } \left ( \frac { 2V } { r } + 2 \pi r^2 \right ) \\ &= \frac { -2V } { r^2 } + 4 \pi r \\ \end{align*} {/eq}

The condition of maxima and minima of area A is found out by equating {eq}\frac { dA } { dr } {/eq} to zero.

{eq}\begin{align*} \frac { dA } { dr } &= 0 \\ \frac { -2V } { r^2 } + 4 \pi r &= 0 \\ r^3 &= \frac { 2V } { 4 \pi } \\ r &= \left ( \frac { 2V } { 4 \pi } \right )^ {\left (\frac { 1 } { 3 } \right ) } \\ \end{align*} {/eq}

The sign of {eq}\begin{align*} \frac { d^2 A } { dr^2 } \end{align*} {/eq} at the above found out value of {eq}r {/eq} tells us whether at this r , the area will be maximum or minimum.

If {eq}\left [ \frac { d^2 A } { dr^2 } \right ]_{\left ( \frac { 2V } { 4 \pi } \right )^ {\left (\frac { 1 } { 3 } \right ) }} < 0 {/eq}, then {eq}r {/eq} is a point of maxima,

If {eq}\left [ \frac { d^2 A } { dr^2 } \right ]_{\left ( \frac { 2V } { 4 \pi } \right )^ {\left (\frac { 1 } { 3 } \right ) }} > 0 {/eq}, then {eq}r {/eq} is a point of minima.

Let us find {eq}\begin{align*} \frac { d^2 A } { dr^2 } \end{align*} {/eq} at {eq}\begin{align*} r = \left ( \frac { 2V } { 4 \pi } \right )^ {\left (\frac { 1 } { 3 } \right ) } \end{align*} {/eq} is calculated as follows;

{eq}\begin{align*} \frac { d^2 A } { dr^2 } &= \frac { d } { dr } \left ( \frac { -2V } { r^2 } + 4 \pi r \right ) \\ &= \frac { 4V } { r^3 } + 4 \pi \\ &= \frac { 4V } { \frac { 2V } { 4\pi } } + 4 \pi \\ &= 12 \pi > 0. \end{align*} {/eq}

Hence, {eq}\begin{align*} r = \left ( \frac { 2V } { 4 \pi } \right )^ {\left (\frac { 1 } { 3 } \right ) } \end{align*} {/eq} is a point of minimum value of the area A.

Substituting {eq}V = 100 \ cm^3 {/eq} in the above expression as follows;

{eq}\begin{align*} r &= \left ( \frac { 2 \times 100 } { 4 \pi } \right )^ {\left (\frac { 1 } { 3 } \right ) } \\ &= 2.515 \ cm. \end{align*} {/eq}

Calculating the height using the formula {eq}h = \frac { V } { \pi r^2 } {/eq} ;

{eq}\begin{align*} h &= \frac { V } { \pi r^2 } \\ &= \frac { 100 } { \pi \times 2.515^2 } \\ &= 5.03 \ cm. \end{align*} {/eq}

Thus, the radius and height of the cylinder are {eq}2.515 \ cm {/eq} and {eq}5.03 \ cm {/eq} respectively. 