Question

The volume of the right circular cylinder is 215600 what should be the radius of the base so that the total surface area is maximum.

Answer 1

Hint:
Firstly, find the value of h from the equation of volume of cylinder and substitute h in the equation of total surface area of the cylinder.
Then, differentiate both sides of the equation thus formed and find the value of radius by making its derivative equal to 0.
Thus, verify by doing double differentiation of S and it must be greater than 0.

Complete step by step solution: Let r and h be radius and height.
Volume of cylinder \[ = \pi {r^2}h\]
Here, \[\pi {r^2}h = 215600\] [given]
\[ \Rightarrow h = \dfrac{{215600}}{{\pi {r^2}}} \ldots \left[ 1 \right]\]
Let the total surface area of a cylinder be S.
\[s = \left( {2\pi {r^2} + 2\pi rh} \right)\]
   \[ = \left( {2\pi {r^2} + 2\pi r\dfrac{{215600}}{{\pi {r^2}}}} \right)\]
  \[ = \left( {2\pi {r^2} + \dfrac{{431200}}{r}} \right)\]
Differentiate with respect to r
\[\dfrac{{ds}}{{dr}} = \left( {4\pi r - \dfrac{{431200}}{{{r^2}}}} \right)\]
Now for maximum take \[\dfrac{{ds}}{{dr}}\]\[ = 0\]
\[ \Rightarrow \]\[4\pi r - \dfrac{{431200}}{{{r^2}}}\]\[ = 0\]
\[ \Rightarrow \dfrac{{4\pi {r^2} - 431200}}{{{r^2}}} = 0\]
\[ \Rightarrow 4\pi {r^2} = 431200\]
\[ \Rightarrow {r^2} = \dfrac{{431200}}{{4\pi }}\]
\[ \Rightarrow r = \sqrt {\dfrac{{431200}}{{4\pi }}} \]
\[ \Rightarrow r = \sqrt {\dfrac{{431200 \times 7}}{{4 \times 22}}} \]
\[ \Rightarrow r = \sqrt {34300} \]
\[ \Rightarrow r = 70\sqrt 7 \]units

The radius of the base so that the total surface area is maximum is \[70\sqrt 7 \]units.

Note:
Second Derivative test:
The second derivative is used to determine the local extremum of a function under given conditions.
If a function has a critical point for which $f'\left( x \right) = 0$ and if the second derivative is positive i.e. its value is greater than 0, then that point is the local maximum of the function. However, if the second derivative at that point is negative i.e. its value is smaller than 0, then that point is the local minimum of the function.