A given quantity of metal is to be cast into a half cylinder with…

Question

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

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Answer

Let h be the height of the cylinder and r be the radius of the semicircular of the ends.
Let V be the volume of the half, therefore
$\text{V}=\frac{1}{2}\pi\text{r}^{2}\text{h}\ ...(\text{i})$
here we have V is constant.
S = area of the rectangular base + area of the two semi circular ends + area of the curved
$\text{S}=\text{h}^{*}2\text{r}+2^{*}\frac{1}{2}\pi\text{r}^{2}+\frac{1}{2}^{*}2\pi\text{rh} $
$\text{S}=2\pi\text{h}=+\pi\text{r}^{2}+\pi\text{rh}\ ...(\text{ii})$
Since V is constant, the values of h terms of V from
$\text{h}=\frac{2\text{r}}{\pi\text{r}^{2}}$
$\text{S}=2\text{r}^{*}\frac{2\text{V}}{\pi\text{r}^{2}}+\pi\text{r}^{2}+\pi\text{r}\frac{2\text{V}}{\pi\text{r}^{2}}$
$\text{S}=\frac{4\text{V}}{\pi\text{r}}+\pi\text{r}^{2}+\frac{2\text{V}}{\text{r}}\ ..(\text{iii})$
Differntiating eq.(iii)
$\frac{\text{dS}}{\text{dr}}=\frac{4\text{V}}{\pi}\Big(-\frac{1}{\text{r}^{2}}\Big)+2\pi\text{r}+{2\text{V}}\Big(\frac{-1}{\text{r}^{2}}\Big)$
$\frac{\text{dS}}{\text{dr}}=2\pi\text{r}-\frac{1}{\text{r}^{2}}\Big(\frac{4\text{V}}{\pi}+2\text{V}\Big) \ ...(\text{iv})$
Now, differentiating (iv)
$\frac{\text{d}^{2}{\text{S}}}{\text{dr}^{2}}=2\pi\text{r}-\Big(\frac{4\text{V}}{\pi}+2\text{V}\Big)\Big(-\frac{2}{\text{r}^{3}}\Big)$
$\frac{\text{dS}}{\text{dr}}=2\pi+\frac{2}{\text{r}^{3}}\Big(\frac{4\text{V}}{\pi}+2\text{V}\Big)\ ...(\text{v})$
for S to be minimum $\frac{\text{ds}}{\text{dr}}=0$
$2\pi\text{r}-\frac{1}{\text{r}^{2}}\Big(\frac{4\text{V}}{\pi}+2\text{V}\Big)=0$
$2\pi\text{r}=\frac{1}{\text{r}^{2}}\Big(\frac{4\text{V}}{\pi}+2\text{V}\Big)$
Then,
$\text{r}^{3}=\frac{1}{2\pi}\Big(\frac{4\text{V}}{\pi}+2\text{V}\Big)=\frac{2\text{V}}{\pi^{2}}+\frac{\text{V}}{\pi}$
Is positive, this for this value S is minimum therefore,
$\text{r}^{3}=\frac{\text{V}}{\pi^{2}}(2+\pi)$
$\text{r}^{3}=\frac{1}{\pi^{2}}^{*}\frac{1}{2}\pi\text{r}^{2}\text{h}(2+\pi)$
$\text{r}=\frac{\text{h}}{2\pi}(2+\pi)$
$\Rightarrow \frac{\text{h}}{2\text{r}}=\frac{\pi}{\pi+2}$
Which is the required ratio of the length half dimametre of its semi circular ends.