Minimise metal for a cylinder with a hemispherical cap

The shared source “Thinking Process” first identifies the goal: find a minimum area, locate stationary values and test their nature. The diagram has two variables, radius r and cylindrical height h. Use fixed volume to eliminate h before differentiating with respect to r; then present the volume relation, area formula, stationary calculation and minimum test in that logical order.

$\pi r^2h+\frac23\pi r^3=100$, so $h=\dfrac{100}{\pi r^2}-\dfrac{2r}{3}$. The exterior metal consists of cylinder side $2\pi rh$, curved hemisphere $2\pi r^2$ and one flat circular base $\pi r^2$. Hence $A=2\pi rh+3\pi r^2=\dfrac{200}{r}+\dfrac53\pi r^2$. Do not count the internal joining circle as a second exposed base.

$dA/dr=-200/r^2+(10/3)\pi r$. Setting this equal to zero gives $r^3=60/\pi$, hence $r=\sqrt[3]{60/\pi}\text{ cm}$. This is the positive radius compatible with the physical model.

The Teacher gives $d^2A/dr^2=400/r^3+(10/3)\pi>0$, so the stationary area is a minimum. The shared source plan also permits a first derivative test. Independently, $r>0$ and $h\ge0$ imply $r^3\le150/\pi$; the candidate is interior and gives $h=r>0$. The area is strictly convex throughout this feasible interval, establishing the global minimum. A positive second derivative is sufficient here, not a necessary test for every minimum.

At $r^3=60/\pi$, $A=\dfrac1r(200+\frac53\pi r^3)=300/r=300\sqrt[3]{\pi/60}\text{ cm}^2$, matching the Teacher answer. The Student copy prints the full planning guidance but leaves the algebraic solution blank. The shared final prompt encourages asking what is needed and how to obtain it; the sequence above follows that advice.