Question

A given quantity of metal sheet is to be cast into a solid half circular cylinder with rectangular base and semicircular ends. Show that in order that the total surface area may be minimum, the ratio of length of cylinder of diameter of its circular ends is $\pi :\pi + 2$

Answer 1

Hint: To check if surface area is minimum or not, you first need to find critical points. And we check if the point is positive or negative.

Complete Step by Step solution:
It is given that a metal sheet is casted into a solid half circular cylinder.
That means, the area of the sheet is equal to the volume of the cylinder.
Let the area of the sheet be A (constant), as it is given and the volume of cylinders is V.
Then, $V = \dfrac{1}{2} \times $Volume of cylinder.
$ = \dfrac{1}{2} \times \pi {r^2}h$
Where, $r$ is the radius of the circular part of the cylinder and $h$is the height of the cylinder.
Total surface area of cylinder will be the sum of $\dfrac{1}{2} \times $curved surface area of cylinder, area of two semicircles, and the area of the rectangular base . . . . . (1)
Curved surface area of cylinder$ = 2\pi rh$ . . . . . . (2)
Area of semicircle$ = \dfrac{{\pi {r^2}}}{2}$ . . . . . . (3)
Area of rectangle$ = 2r \times h$ . . . . . . . (4)
Since in this case the height of the cylinder is the length of the rectangular base and the diameter of the semicircle is the breadth of the rectangular base.
Putting the values from equations (2), (3) and (4) into equation (1) we get
Total surface area of the cylinder,
$S = \dfrac{1}{2} \times 2\pi rh + 2 \times \dfrac{{\pi {r^2}}}{2} + 2r \times h$
$S = \pi rh + \pi {r^2} + 2rh$ . . . . . . (5)
Since, area of sheet, B equal to the volume of cylinder, we have $\dfrac{1}{2} \times \pi {r^2}h = A$
$h = \dfrac{{2A}}{{\pi {r^2}}}$ . . . . . (6)
Put this value in (5), to get
$S = \pi r \times \dfrac{{2A}}{{\pi {r^2}}} + \pi {r^2} + 2r \times \dfrac{{2A}}{{\pi {r^2}}}$
$ = \dfrac{{2A}}{r} + \pi {r^2} + \dfrac{{4A}}{{\pi r}}$ . . . . . . (7)
For the surface area to be minimum or maximum$\dfrac{{ds}}{{dr}} = 0$.
By differentiating equations (7) we get
$\dfrac{{ds}}{{dr}} = 2A\dfrac{{( - 1)}}{{{r^2}}} + 2\pi r + \dfrac{{4A( - 1)}}{{\pi {r^2}}}$
$\dfrac{{ds}}{{dr}} = \dfrac{{2A}}{{{r^2}}} + 2\pi r - \dfrac{{4A}}{{\pi {r^2}}}$ $\left( {\because \dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}} \right)$
$\therefore \dfrac{{ds}}{{dr}} = 0$
$\dfrac{{ - 2A}}{{{r^2}}} + 2\pi r - \dfrac{{4A}}{{\pi {r^2}}} = 0$
Multiplying both the sides by $\pi {r^2},$we get
$ \Rightarrow - 2\pi A + 2{\pi ^2}{r^3} - 4A = 0$
$ \Rightarrow 2{\pi ^2}{r^3} = 4A + 2\pi A$
$ = (2 + \pi )2A$
$ \Rightarrow {r^3} = \dfrac{{(\pi + 2)2A}}{{2{\pi ^2}}}$ . . . . . (8)
Now, we need to check if the total surface area will be minimum for this value of $r$.
For that differentiate$\dfrac{{ds}}{{dr}}$
$\dfrac{{{d^2}s}}{{d{r^2}}} = \dfrac{{4A}}{{{r^3}}} + 2\pi + \dfrac{{8A}}{{\pi {r^3}}}$
${\left( {\dfrac{{{d^2}s}}{{d{r^2}}}} \right)_{{r^3} = \dfrac{{(\pi + 2)A}}{{{\pi ^2}}}}} = \dfrac{{4A{\pi ^2}}}{{(\pi + 2)}} + 2\pi + \dfrac{{8A{\pi ^2}}}{{\pi (\pi + 2)A}}$
$ = \dfrac{{4{\pi ^2}}}{{\pi + 2}} + 2\pi + \dfrac{{8\pi }}{{\pi + 2}}$
Clearly, all the values are positive.
$\therefore {\left( {\dfrac{{{d^2}s}}{{d{r^2}}}} \right)_r} > 0.$
$ \Rightarrow $S is minimum.
$\because $for minima at point a, ${\left( {\dfrac{{{d^2}s}}{{d{r^2}}}} \right)_a} > 0$
Now, we need to find the ratio of length of the cylinder to the diameter of its circular ends i.e.$\dfrac{h}{{2r}}.//$
$ \Rightarrow \dfrac{h}{{2r}} = \dfrac{1}{{2r}} \times \dfrac{{2A}}{{\pi {r^2}}}$
$ = \dfrac{A}{{\pi {r^3}}} $
$ = \dfrac{A}{{\pi \times \dfrac{{(\pi + 2)A}}{\pi }}}$
$ \Rightarrow \dfrac{h}{{2r}} = \dfrac{\pi }{{(\pi + 2)}}$
Hence, it is proved that the ratio of the length of the cylinder to the diameter of its circular ends is$\pi :(\pi + 2)$

Note: Equation of the question will most probably be in two variables. You have to make sure to find a relation between the variables. So that you could simplify equations in a single variable. Just like in this example, we found the relation between $h$and $r,$ and converted the equation of S in terms of
$r$.