(a) Express $h$ via the constraint A1
$V_{0} = \pi r^{2} h \;\Longrightarrow\; h = \dfrac{V_{0}}{\pi r^{2}}$.
(b) Surface area formula M1·A1
Open-top: base disc $+$ curved lateral surface, no top. So
$$ S \;=\; \pi r^{2} + 2\pi r h \;=\; \pi r^{2} + 2\pi r \cdot \frac{V_{0}}{\pi r^{2}} \;=\; \pi r^{2} + \frac{2 V_{0}}{r}. \;\;\text{AG} $$
(c) Differentiate and find critical radius M1·A1·A1·A1
$$ \frac{dS}{dr} \;=\; 2\pi r - \frac{2 V_{0}}{r^{2}}. $$
Set $\dfrac{dS}{dr} = 0$:
$$ 2\pi r \;=\; \frac{2 V_{0}}{r^{2}} \;\Longrightarrow\; \pi r^{3} \;=\; V_{0} \;\Longrightarrow\; r^{3} \;=\; \frac{V_{0}}{\pi} \;\Longrightarrow\; r^{*} \;=\; \left(\frac{V_{0}}{\pi}\right)^{1/3}. \;\;\text{AG} $$
(d) Show $h^{*} = r^{*}$ M1·A1
Using $h = V_{0}/(\pi r^{2})$ at $r = r^{*}$:
$$ h^{*} \;=\; \frac{V_{0}}{\pi (r^{*})^{2}} \;=\; \frac{V_{0}}{\pi} \cdot \frac{1}{(r^{*})^{2}}. $$
But $(r^{*})^{3} = V_{0}/\pi$, so $V_{0}/\pi = (r^{*})^{3}$. Therefore
$$ h^{*} \;=\; \frac{(r^{*})^{3}}{(r^{*})^{2}} \;=\; r^{*}. \;\;\text{AG} $$
(e) Second-derivative test M1·A1
Differentiate again:
$$ \frac{d^{2}S}{dr^{2}} \;=\; 2\pi + \frac{4 V_{0}}{r^{3}}. $$
For all $r > 0$ this is positive (sum of two positive terms), in particular at $r = r^{*}$. So $S$ is concave up at the critical point, hence $r^{*}$ is a
local minimum. (Since it is the only critical point on $r > 0$ and $S \to \infty$ as $r \to 0^{+}$ and as $r \to \infty$, it is the global minimum on the domain.)
(f) Numerical evaluation at $V_{0} = 1000$ M1·A1
$r^{*} = (1000/\pi)^{1/3} = (318.31\ldots)^{1/3} \approx 6.8278\ldots \approx 6.83$ cm. $\;$ $h^{*} = r^{*} \approx 6.83$ cm.
$$ S(r^{*}) \;=\; \pi (r^{*})^{2} + \frac{2 \cdot 1000}{r^{*}} \;\approx\; \pi (6.8278)^{2} + \frac{2000}{6.8278} \;\approx\; 146.43 + 292.85 \;\approx\; 439.3 \;\approx\; 440 \text{ cm}^{2}. $$
All three to 3 sf.
"$h = r$ for open-top, $h = 2r$ for closed" — the canonical can-design rule. The open-top result $h^{*} = r^{*}$ here contrasts with the closed-can result $h^{*} = 2r^{*}$ (which you get by setting up $S = 2\pi r^{2} + 2\pi r h$ and minimising under fixed volume). The geometric reason: a top adds another $\pi r^{2}$ to surface area, doubling the "disc penalty" relative to the lateral. The lateral surface wants $r$ small (less circumference) while the disc(s) want $r$ small (less area), but the volume constraint $\pi r^{2} h = V_{0}$ couples them. Calculus picks the unique compromise. This same template — "express one variable via the constraint, substitute into the objective, differentiate, set to zero, verify with second derivative" — runs through every IB optimisation question in Paper 2. Master the 5-step recipe and the rest is arithmetic.
(a) 由约束写出 $h$ A1
$V_{0} = \pi r^{2} h \;\Longrightarrow\; h = \dfrac{V_{0}}{\pi r^{2}}$。
(b) 表面积公式 M1·A1
无盖：底圆盘 $+$ 侧面曲面，无顶。故
$$ S \;=\; \pi r^{2} + 2\pi r h \;=\; \pi r^{2} + 2\pi r \cdot \frac{V_{0}}{\pi r^{2}} \;=\; \pi r^{2} + \frac{2 V_{0}}{r}. \;\;\text{AG} $$
(c) 求导并取临界半径 M1·A1·A1·A1
$$ \frac{dS}{dr} \;=\; 2\pi r - \frac{2 V_{0}}{r^{2}}. $$
令 $\dfrac{dS}{dr} = 0$：
$$ 2\pi r \;=\; \frac{2 V_{0}}{r^{2}} \;\Longrightarrow\; \pi r^{3} \;=\; V_{0} \;\Longrightarrow\; r^{3} \;=\; \frac{V_{0}}{\pi} \;\Longrightarrow\; r^{*} \;=\; \left(\frac{V_{0}}{\pi}\right)^{1/3}. \;\;\text{AG} $$
(d) 证 $h^{*} = r^{*}$ M1·A1
在 $r = r^{*}$ 处用 $h = V_{0}/(\pi r^{2})$：
$$ h^{*} \;=\; \frac{V_{0}}{\pi (r^{*})^{2}}. $$
由 $(r^{*})^{3} = V_{0}/\pi$，即 $V_{0}/\pi = (r^{*})^{3}$。故
$$ h^{*} \;=\; \frac{(r^{*})^{3}}{(r^{*})^{2}} \;=\; r^{*}. \;\;\text{AG} $$
(e) 二阶导检验 M1·A1
再次求导：
$$ \frac{d^{2}S}{dr^{2}} \;=\; 2\pi + \frac{4 V_{0}}{r^{3}}. $$
对所有 $r > 0$ 该值为正（两正数之和），尤其在 $r = r^{*}$ 处为正。故 $S$ 在临界点处凹向上，$r^{*}$ 为
局部极小。（在 $r > 0$ 上仅此一临界点，且 $r \to 0^{+}$ 与 $r \to \infty$ 时 $S \to \infty$，故为全局极小。）
(f) $V_{0} = 1000$ 时的数值 M1·A1
$r^{*} = (1000/\pi)^{1/3} = (318.31\ldots)^{1/3} \approx 6.8278\ldots \approx 6.83$ cm。$\;$ $h^{*} = r^{*} \approx 6.83$ cm。
$$ S(r^{*}) \;=\; \pi (r^{*})^{2} + \frac{2 \cdot 1000}{r^{*}} \;\approx\; \pi (6.8278)^{2} + \frac{2000}{6.8278} \;\approx\; 146.43 + 292.85 \;\approx\; 439.3 \;\approx\; 440 \text{ cm}^{2}. $$
均取 3 位有效数字。
"无盖 $h = r$，封闭 $h = 2r$" — 罐子设计的标准结论。本题无盖结果 $h^{*} = r^{*}$ 与封闭罐结果 $h^{*} = 2r^{*}$（设 $S = 2\pi r^{2} + 2\pi r h$ 同法极小化）形成对照。几何理由：多一个顶盖再加 $\pi r^{2}$，使"圆盘惩罚"相对侧面翻倍。侧面欲 $r$ 小（周长小）、圆盘也欲 $r$ 小（面积小），但体积约束 $\pi r^{2} h = V_{0}$ 把二者绑定；微积分挑出唯一折中。同一模板 — "由约束表一变量，代回目标，求导取零，二阶导验证" — 贯穿 IB Paper 2 所有优化题。掌握这 5 步流程，剩下只是算术。