Volumes Practice Test 3

Here, V = $\mathit{\pi }{\mathit{r}}^2\mathit{h}$
∴ h = $\frac{\mathit{V}}{\mathit{\pi }{\mathit{r}}^2}$
and also, S = 2πrh + $2\mathit{\pi }{\mathit{r}}^2$ = 2πr × $\frac{\mathit{V}}{\mathit{\pi }{\mathit{r}}^2}+2\mathit{\pi }{\mathit{r}}^2$
= $\frac{2\mathit{V}}{\mathit{r}}+2\mathit{\pi }{\mathit{r}}^2$
Since, $\frac{\mathit{d}\mathit{S}}{\mathit{d}\mathit{r}}$ = $-\frac{2\mathit{V}}{{\mathit{r}}^2}+4\mathit{\pi }\mathit{r}$ and also, $\frac{{\mathit{d}}^2\mathit{S}}{\mathit{d}\mathit{r}}$ = $\left(\frac{4\mathit{V}}{{\mathit{r}}^3}+4\mathit{\pi }\right)$ > 0
Hence, S is minimum when $\frac{\mathit{d}\mathit{S}}{\mathit{d}\mathit{r}}$ = 0
Then, $-\frac{2\mathit{V}}{{\mathit{r}}^2}+4\mathit{\pi }\mathit{r}$ = 0
⇒ $-\frac{2\mathit{\pi }{\mathit{r}}^2\mathit{h}}{{\mathit{r}}^2}+4\mathit{\pi }\mathit{r}$ = 0
h = 2r = d
Since, height is equal to its diameter.