Question

Show that the height of a closed right circular cylinder of given surface and maximum volume, is equal to the diameter of its base.

Answer 1

In this question we have been given a closed right circular cylinder, whose surface is given and it has maximum volume. We need to show that the height of this cylinder is equal to the diameter of its base. So, for that firstly we will let the surface area, volume, radius and height as any of the variables. After that we have given the surface area whose formula is $S = 2\pi {r^2} + 2\pi rh$, from here we will find the relation between height and radius keeping the surface area constant as it is given. Then we need to maximise the volume using the above relation, for which the double derivative must be negative. After that we will find the value of height in terms of radius using the relation $\dfrac{{dV}}{{dr}} = 0$.

Complete step by step answer:

We have been provided with a closed right circular cylinder,
Let the surface area of the cylinder be S, volume of the cylinder be V, radius of the cylinder be r, diameter of the base of the cylinder be d and height of the cylinder be h.
So, the surface area of the cylinder is given whose formula is: $S = 2\pi {r^2} + 2\pi rh$.
Now we will be using this formula to find the relation between h and r,
So, the relation would be:$h = \dfrac{{S - 2\pi {r^2}}}{{2\pi r}}$
Also, we need to maximise the volume of this cylinder so, for that firstly the formula of volume of cylinder is:$V = \pi {r^2}h$
Now we will keep the value of h which we found from the given surface area in terms of r,
So now the volume of the cylinder will become:
$\begin{gathered} V = \pi {r^2}\left[ {\dfrac{S}{{2\pi }}\left( {\dfrac{1}{r}} \right) - r} \right] \\\ = > \dfrac{{Sr}}{2} - \pi {r^3} \\\ \end{gathered}$
So, for maximum volume the double derivative must be negative,
So, differentiating dv with respect to dr,
$\dfrac{{dV}}{{dr}} = \dfrac{S}{2} - 3\pi {r^2}$
Now differentiating again,
$\dfrac{{{d^2}V}}{{d{r^2}}} = 0 - 2(3\pi r)$ using the formula $({a^n}dn = n{a^{n - 1}}dn)$
$\dfrac{{{d^2}V}}{{d{r^2}}} = - 6\pi r$
So, the double derivative is negative which means the volume of the cylinder is maximum,
Now, $\dfrac{{dV}}{{dr}} = 0$
So, $\dfrac{S}{2} - 3\pi {r^2} = 0$
Now we will simplify the above equation further,
$\dfrac{S}{2} = 3\pi {r^2}$
Now we will find the value of S, which comes out to be:$S = 6\pi {r^2}$
Now we will keep this value in formula of surface area to find out the relation between h and r,
So, now the equation would become:$2\pi {r^2} + 2\pi rh = 6\pi {r^2}$
Solving the equation further, $2\pi {r^2} + 2\pi rh - 6\pi {r^2} = 0$
Now, simplifying the equation: $- 4\pi {r^2} + 2\pi rh = 0$
Now solving the equation further, we will get:$- 4\pi {r^2} = - 2\pi h$
So, the relation between h and r would be:$h = 2r$, which is the diameter of the base of the cylinder.
so, this is the solution to this question.

Note:
In this question we must refer to the figure for better understanding, you can also use the first derivative test to show the volume is maximum, but second would be preferred more as chances of mistakes reduces in it, but derivative test is important to show if the volume is maximum or not.