一个底面半径为 $r$ 且高为 $h$ 的封闭直圆锥，其固定总表面积为 $k\pi$ 单位$^2$，其中 $k$ 是正常数。证明 $r^2 = \frac{k^2}{(h^2 + 2k)}$。已知 $r$ 和 $h$ 是变化的，求圆锥的最大可能体积，结果用关于 $k$ 的精确形式表示。 [底面半径为 $r$ 且高为 $h$ 的圆锥体积公式为 $V = \frac{1}{3}\pi r^2 h$，圆锥的侧面积公式为 $A = \pi r l$，其中 $l$ 为斜高。]

1. Set up the equation for the total surface area and simplify.

$\pi r^2 + \pi r l = k\pi$
$r^2 + r\sqrt{r^2 + h^2} = k$

2. Rearrange the equation and square both sides to eliminate the square root.

$r\sqrt{r^2 + h^2} = k - r^2$
$(r\sqrt{r^2 + h^2})^2 = (k - r^2)^2$
$r^2(r^2 + h^2) = k^2 - 2kr^2 + r^4$
$r^4 + r^2 h^2 = k^2 - 2kr^2 + r^4$

3. Solve for $r^2$ to obtain the required identity.

$r^2 h^2 = k^2 - 2kr^2$
$r^2 h^2 + 2kr^2 = k^2$
$r^2(h^2 + 2k) = k^2$
$r^2 = \frac{k^2}{h^2 + 2k} \quad \text{(shown)}$

4. Express the volume of the cone in terms of a single variable $h$.

$V = \frac{1}{3}\pi r^2 h$
$V = \frac{1}{3}\pi \left(\frac{k^2}{h^2 + 2k}\right) h = \frac{\pi k^2 h}{3(h^2 + 2k)}$

5. Differentiate the volume $V$ with respect to $h$ using the quotient rule.

$\frac{dV}{dh} = \frac{\pi k^2}{3} \left[ \frac{(h^2 + 2k)(1) - h(2h)}{(h^2 + 2k)^2} \right]$
$\frac{dV}{dh} = \frac{\pi k^2}{3} \left[ \frac{h^2 + 2k - 2h^2}{(h^2 + 2k)^2} \right] = \frac{\pi k^2(2k - h^2)}{3(h^2 + 2k)^2}$

6. Set the derivative to zero to find the stationary point and verify it gives a maximum.

For a stationary point, $\frac{dV}{dh} = 0$:
$2k - h^2 = 0 \implies h^2 = 2k$
$h = \sqrt{2k} \quad \text{(since } h > 0\text{)}$
Using the first derivative test:

• For $0 < h < \sqrt{2k}$, $2k - h^2 > 0 \implies \frac{dV}{dh} > 0$
• For $h > \sqrt{2k}$, $2k - h^2 < 0 \implies \frac{dV}{dh} < 0$
  Thus, $V$ reaches a maximum when $h = \sqrt{2k}$.

7. Substitute $h = \sqrt{2k}$ into the volume expression to calculate the greatest possible volume.

$V_{\text{max}} = \frac{\pi k^2 (\sqrt{2k})}{3((\sqrt{2k})^2 + 2k)}$
$V_{\text{max}} = \frac{\pi k^2 \sqrt{2k}}{3(2k + 2k)} = \frac{\pi k^2 \sqrt{2k}}{12k} = \frac{\pi k \sqrt{2k}}{12}$