Show that the right circular cylinder of given surface area and maximum volume is such that its height is equal to the diameter of the base.

(N/A) Let $r$ and $h$ be the radius and height of the cylinder respectively.
The surface area $S$ of the cylinder is given by $S = 2\pi r^2 + 2\pi rh$.
From this,we can express $h$ in terms of $r$ and $S$:
$h = \frac{S - 2\pi r^2}{2\pi r} = \frac{S}{2\pi r} - r$.
The volume $V$ of the cylinder is $V = \pi r^2 h$.
Substituting the expression for $h$:
$V = \pi r^2 \left( \frac{S}{2\pi r} - r \right) = \frac{Sr}{2} - \pi r^3$.
To find the maximum volume,we differentiate $V$ with respect to $r$:
$\frac{dV}{dr} = \frac{S}{2} - 3\pi r^2$.
Setting $\frac{dV}{dr} = 0$ gives $\frac{S}{2} = 3\pi r^2$,so $S = 6\pi r^2$.
Now,find the second derivative to check for maximum:
$\frac{d^2V}{dr^2} = -6\pi r$.
Since $r > 0$,$\frac{d^2V}{dr^2} < 0$,which confirms that the volume is maximum at $S = 6\pi r^2$.
Substituting $S = 6\pi r^2$ into the expression for $h$:
$h = \frac{6\pi r^2 - 2\pi r^2}{2\pi r} = \frac{4\pi r^2}{2\pi r} = 2r$.
Since $2r$ is the diameter of the base,the height of the cylinder is equal to its diameter.