[底面半径为 $r$ 且高为 $h$ 的圆锥的体积由 $V = \frac{1}{3}\pi r^2 h$ 给出。] 一位制造商用同一种材料制作了一个漏斗状的装饰品，它由两部分组成，如图 1所示。

• 一个半径为 $r$ cm、高为 $h$ cm 且母线长为 4 cm 的直圆锥，
• 一个半径为 $ar$ cm 且高为 $r$ cm 的圆柱，其中 $0 < a < 1$。

如图 2所示，从原始圆锥的顶点处切除一个半径为 $ar$ cm 的相似圆锥。圆锥的剩余部分与圆柱拼接在一起，形成如图 3所示的漏斗。可以假设漏斗的厚度忽略不计。 已知该装饰品的体积为 $V\text{ cm}^3$，求用 $a$ 和 $r$ 表示 $V$ 的表达式。

1. Express the total volume $V$ in terms of $r$ by evaluating the volumes of the frustum and cylinder.

From the given information, the slant height of the original cone is $4$. Let its vertical height be $h$. By Pythagoras' Theorem:
$r^2 + h^2 = 4^2 \implies h = \sqrt{16 - r^2}$

The similar cone removed from the vertex has radius $ar$ and, by similarity, height $ah$.
Volume of the frustum = $\frac{1}{3}\pi r^2 h - \frac{1}{3}\pi (ar)^2 (ah) = \frac{1}{3}\pi r^2 h (1 - a^3)$
Volume of the cylinder = $\pi (ar)^2(r) = \pi a^2 r^3$

Total volume $V$:
$V = \frac{1}{3}\pi r^2 \sqrt{16 - r^2} (1 - a^3) + \pi a^2 r^3$

2. Substitute $a = 0.25$ to express $V$ solely in terms of $r$.

Given $a = 0.25 = \frac{1}{4}$:
$a^2 = \frac{1}{16}$
$a^3 = \frac{1}{64} \implies 1 - a^3 = \frac{63}{64}$

Substitute into $V$:
$V = \frac{1}{3}\pi r^2 \sqrt{16 - r^2} \left(\frac{63}{64}\right) + \pi \left(\frac{1}{16}\right) r^3$
$V = \frac{21\pi}{64} r^2 \sqrt{16 - r^2} + \frac{\pi}{16} r^3$
$V = \frac{\pi}{64} \left[ 21r^2 \sqrt{16 - r^2} + 4r^3 \right]$

3. Differentiate $V$ with respect to $r$ using the product and chain rules.

$\frac{dV}{dr} = \frac{\pi}{64} \left[ \frac{d}{dr}\left(21r^2 (16 - r^2)^{1/2}\right) + \frac{d}{dr}\left(4r^3\right) \right]$

Applying the product rule on the first term:
$\frac{dV}{dr} = \frac{\pi}{64} \left[ 42r (16 - r^2)^{1/2} + 21r^2 \left(\frac{1}{2}\right)(16 - r^2)^{-1/2}(-2r) + 12r^2 \right]$
$\frac{dV}{dr} = \frac{\pi}{64} \left[ 42r \sqrt{16 - r^2} - \frac{21r^3}{\sqrt{16 - r^2}} + 12r^2 \right]$

4. Equate $\frac{dV}{dr}$ to 0 for a maximum value and substitute $r = r_1$.

For $V$ to be a maximum, $\frac{dV}{dr} = 0$. Let $r = r_1$.
$\frac{\pi}{64} \left[ 42r_1 \sqrt{16 - r_1^2} - \frac{21r_1^3}{\sqrt{16 - r_1^2}} + 12r_1^2 \right] = 0$

Since $r_1 > 0$, divide the entire equation by $3r_1$:
$14 \sqrt{16 - r_1^2} - \frac{7r_1^2}{\sqrt{16 - r_1^2}} + 4r_1 = 0$

Multiply through by $\sqrt{16 - r_1^2}$ to clear the denominator:
$14(16 - r_1^2) - 7r_1^2 + 4r_1\sqrt{16 - r_1^2} = 0$
$224 - 14r_1^2 - 7r_1^2 + 4r_1\sqrt{16 - r_1^2} = 0$
$224 - 21r_1^2 = -4r_1\sqrt{16 - r_1^2}$

5. Square both sides to eliminate the square root and form the required equation.

Square both sides:
$(224 - 21r_1^2)^2 = (-4r_1\sqrt{16 - r_1^2})^2$
$50176 - 2(224)(21)r_1^2 + 441r_1^4 = 16r_1^2(16 - r_1^2)$
$50176 - 9408r_1^2 + 441r_1^4 = 256r_1^2 - 16r_1^4$

Rearranging the terms to one side:
$(441 + 16)r_1^4 - (9408 + 256)r_1^2 + 50176 = 0$
$457r_1^4 - 9664r_1^2 + 50176 = 0$ (shown)

1. Solve the equation from part (a) to find the possible positive roots for $r$.

$457r^4 - 9664r^2 + 50176 = 0$
Let $y = r^2$. Using the quadratic formula:
$r^2 = \frac{9664 \pm \sqrt{(-9664)^2 - 4(457)(50176)}}{2(457)}$
$r^2 = \frac{9664 \pm \sqrt{1671168}}{914}$
$r^2 = \frac{4832 \pm 64\sqrt{102}}{457}$
Evaluating these gives:
$r^2 \approx 9.1589 \quad \text{or} \quad r^2 \approx 11.9877$
Taking the positive square roots (since radius $r > 0$):
$r \approx 3.026 \quad \text{or} \quad r \approx 3.462$

2. Test the first positive root in the original, un-squared condition for a stationary point to show it is extraneous.

From part (a), equating $\frac{dV}{dr}$ to zero leads to the condition before squaring:
$21r^2 - 224 = 4r\sqrt{16 - r^2}$
Substitute $r \approx 3.026$ ($r^2 \approx 9.1589$) into both sides:
$\text{LHS} = 21(9.1589) - 224 \approx -31.66$
$\text{RHS} = 4(3.026)\sqrt{16 - 9.1589} = 12.104(2.6155) \approx 31.66$
Since LHS $\neq$ RHS (they have opposite signs), $r \approx 3.026$ is an extraneous root introduced by squaring and does not give a stationary value of $V$.

3. Find the value of $h$ corresponding to the valid root.

The valid root for the stationary point must be the other one, $r^2 \approx 11.9877$.
Using the relationship between slant height, radius, and height from the original cone ($r^2 + h^2 = 4^2 = 16$):
$h = \sqrt{16 - r^2}$
$h = \sqrt{16 - 11.9877}$
$h = \sqrt{4.0123}$
$h \approx 2.003$
Thus, $h = 2.00$ cm (to 3 s.f.).