A vacuum flask, used to keep drinks hot, is represented by a closed cylinder whose internal radius is $r\,\text{cm}$ and internal height is $h\,\text{cm}$. The flask has volume $1000\,\text{cm}^3$. It is most efficient when the total internal surface area, $A\,\text{cm}^2$, is as small as possible.
(i)[3]
Show that, in terms of $r$, $A = 2\pi r^2 + \frac{2000}{r}$.
(ii)[5]
If $r$ is allowed to change, determine the value of $r$, correct to $1$ decimal place, for which $A$ is stationary, and confirm that the flask is most efficient for this value of $r$.
Worked solution & mark scheme
This 8-mark question has a full step-by-step worked solution and mark scheme. One marking point: “States the surface area formula as $A=2\pi r^2+2\pi rh$” …