# a closed right circular cylinder has volume 2156 cubic units what could be the radius of the base so that the total surface area maximum

Let r and h be the radius and height of the right circular cylinder respectively.
Now, we know that the volume of the cylinder is${\mathrm{\pi r}}^{2}\mathrm{h}$. So,

Next, let the total surface area of the cylinder be S. So,

Differentiating the total surface area with respect to r, we get
$\frac{dS}{dr}=\left(4\mathrm{\pi r}-\frac{4312}{{\mathrm{r}}^{2}}\right)$
Also,
$\frac{{d}^{2}S}{d{r}^{2}}=\left(4\mathrm{\pi }+\frac{8624}{{\mathrm{r}}^{3}}\right)$

Now,
$\frac{dS}{dr}=0\phantom{\rule{0ex}{0ex}}\left(4\mathrm{\pi r}-\frac{4312}{{\mathrm{r}}^{2}}\right)=0\phantom{\rule{0ex}{0ex}}\frac{4{\mathrm{\pi r}}^{3}-4312}{{r}^{2}}=0\phantom{\rule{0ex}{0ex}}$
$4{\mathrm{\pi r}}^{3}=4312\phantom{\rule{0ex}{0ex}}{\mathrm{r}}^{3}=\frac{4312}{4\mathrm{\pi }}\phantom{\rule{0ex}{0ex}}\mathrm{r}=\sqrt{\frac{4312×7}{4×22}}$
$r=\sqrt{343}\phantom{\rule{0ex}{0ex}}r=7$

And
${\left[\frac{{d}^{2}S}{d{r}^{2}}\right]}_{r=7}=\left(4\mathrm{\pi }+\frac{8624}{343}\right)>0$
Therefore, the total surface area of the cylinder is maximum when radius is 7 units.

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