Question

A closed cylinder has volume $2156c{m^3}$ . What will be the radius of it’s base so that its total surface area is minimum?

Answer 1

Hint: In order to solve this question, the basic knowledge about a cylinder and it’s standard formulae is required. A cylinder is a three dimensional solid shape consisting of two parallel circular shaped bases ( always congruent and parallel to each other ) which are joined by a curved surface. The basic formulae for a cylinder having radius r and height h is given by $\left( 1 \right){\text{Surface area of a cylinder = 2}}\pi rh + 2\pi {r^2}$ and $\left( 2 \right){\text{Volume of a cylinder = }}\pi {{\text{r}}^2}h$

Complete step-by-step solution:
The basic diagram for a cylinder is shown below;

                            Figure : Three dimensional representation of a cylinder
                             $\left( {{\text{radius = r and height = h }}} \right)$
According to the question;
Given : Volume of cylinder $ = 2156c{m^3}$
By the formula stated above for the volume of a cylinder, we can equate;
$ \Rightarrow 2156c{m^3} = \pi {r^2}h$
We can calculate the expression for the height of the cylinder from the above equation as;
$ \Rightarrow h = \dfrac{{2156}}{{\pi {r^2}}}{\text{ }}......\left( 1 \right)$
We know that the surface area of cylinder is given by;
$ \Rightarrow S = 2\pi rh + 2\pi {r^2} = 2\pi r\left( {h + r} \right)$
Put the value of h in the above expression from equation $\left( 1 \right)$ , we get;
$ \Rightarrow S = 2\pi r\left( {r + \dfrac{{2156}}{{\pi {r^2}}}} \right)$
Multiplying $2\pi r$ in the inside ;
$ \Rightarrow S = 2\pi {r^2} + \dfrac{{4312}}{r}{\text{ }}......\left( 2 \right)$
Differentiate the above equation with respect to the radius $\left( r \right)$ , we get ;
$ \Rightarrow S' = \dfrac{d}{{dx}}\left( {2\pi {r^2} + \dfrac{{4312}}{r}} \right)$
$\left[ {\because \dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}} \right]$
Using the above standard differentiation rule;
$\therefore \dfrac{d}{{dr}}\left( {\dfrac{1}{r}} \right) = - \dfrac{1}{{{r^2}}}$
$ \Rightarrow S' = 4\pi r - \dfrac{{4312}}{{{r^2}}}{\text{ }}......\left( 3 \right)$
Equating equation $\left( 3 \right)$ with zero , we get ;
$ \Rightarrow 4\pi r - \dfrac{{4312}}{{{r^2}}} = 0$
Further solving the above equation ;
$ \Rightarrow 4\pi r = \dfrac{{4312}}{{{r^2}}}$
Rearranging the above equation;
$ \Rightarrow {r^3} = \dfrac{{4312}}{{4\pi }} = \dfrac{{1078}}{\pi }$
Therefore, we get the value of radius as ;
$ \Rightarrow r = {\left( {\dfrac{{1078}}{\pi }} \right)^{\dfrac{1}{3}}}$
We can further simplify for the value of radius by putting $\pi = \dfrac{{22}}{7}$ in the above equation;
$ \Rightarrow r = {\left( {1078 \times \dfrac{7}{{22}}} \right)^{\dfrac{1}{3}}}$
On further simplification;
$ \Rightarrow r = {\left( {2 \times 7 \times 7 \times 11 \times \dfrac{7}{{2 \times 11}}} \right)^{\dfrac{1}{3}}}$
$ \Rightarrow r = {\left( {{7^3}} \right)^{\dfrac{1}{3}}}$
Therefore, $r = 7cm.$
Differentiating equation $\left( 3 \right)$ with respect to r again , we get ;
$ \Rightarrow S'' = 4\pi - \left( {\dfrac{{4312 \times - 2}}{{{r^3}}}} \right)$
$\because \dfrac{d}{{dr}}\left( { - \dfrac{1}{{{r^2}}}} \right) = \dfrac{2}{{{r^3}}}$
Therefore, we get the equation for the double derivative as ;
$ \Rightarrow S'' = 4\pi + \dfrac{{8624}}{{{r^3}}}{\text{ }}......\left( 4 \right)$
From equation $\left( 4 \right)$ , we can say that ;
$ \Rightarrow S'' > 0$ $\left( {\because \pi = \dfrac{{22}}{7}{\text{ and }}r = 7cm{\text{ ; both are positive values}}} \right)$
Therefore , this is the point of minima with critical point $r = 7cm$ .
Hence, we can say that the radius of the base must be $7cm$ for the surface area of the cylinder to be minimum.
So, the correct answer for this question is $\text{radius} = 7cm$.

Note: To calculate maxima or minima for a given function, there are certain steps which need to be followed: $\left( 1 \right)$ Differentiate the given function with respect to the varying parameter. $\left( 2 \right)$ To calculate maxima or minima equate the first derivative with zero to find out the critical point. $\left( 3 \right)$ Find out the second derivative of the given function. $\left( 4 \right)$ Put the value(s) of the critical point in the second derivative. $\left( 5 \right)$ After putting the value of the critical point, if the value of the second derivative function is greater than zero or positive then that particular point is called point of minima . $\left( 6 \right)$ After putting the value of the critical point, if the value of the second derivative function is less than zero or negative, then that point is called point of maxima for the given function.