[The volume of a cone of base radius $r$ and height $h$ is given by $V = \frac{1}{3}\pi r^2 h$.] A manufacturer makes a funnel-shaped ornament from the same material which consists of two parts as shown in Figure 1.

• a right cone of radius $r$ cm, height $h$ cm and a slant height of 4 cm,
• a cylinder with radius $ar$ cm and height $r$ cm, where $0 < a < 1$.

From the original cone, a similar cone with radius $ar$ cm is removed from the vertex as shown in Figure 2. The remaining part of the cone is joined to the cylinder to form the funnel as shown in Figure 3. It may be assumed that the thickness of the funnel is negligible. Given that the volume of the ornament is $V\text{ cm}^3$, find $V$ in terms of $a$ and $r$.

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.).