In the diagram, point $Q$ is $400\text{ m}$ vertically above point $A$. Initially, a balloon is at point $Q$ and moves horizontally in a straight line at a constant speed of $10\text{ m}$ per second away from an observer stationed at point $A$. At time $t$ seconds, after the start, the balloon is at point $P$ where $QP$ is $x\text{ m}$ and the angle of elevation of the balloon from the observer at $A$ is $\beta$ radians. Find exactly the rate of change of $\beta$ at the instant when $x$ is $100\text{ m}$. A gardener designs a flower bed $ABC$ in the shape of an isosceles triangle inscribed in a circular plot of land with radius $6$ meters, as shown in the diagram. The vertices of the triangle lie on the circumference of the circle and $AB = AC$. The angle $BAC$ is $\theta$, where $\theta$ is acute. Express the area of the flower bed $ABC$ in terms of $\theta$. As $\theta$ varies, use differentiation to find the exact maximum possible area of the flower bed, and show that it is a maximum.

1. Relate the angle of elevation $\beta$ to the horizontal distance $x$.

From the given diagram, the balloon forms a right-angled triangle with the ground. The constant vertical height is $400\text{ m}$ and the horizontal distance is $x\text{ m}$.
$\tan \beta = \frac{400}{x}$

2. Differentiate the equation with respect to time $t$.

Differentiating both sides implicitly with respect to $t$:
$\sec^2 \beta \frac{d\beta}{dt} = -\frac{400}{x^2} \frac{dx}{dt}$

3. Determine the exact value of $\sec^2 \beta$ at the specified instant.

When $x = 100$,
$\tan \beta = \frac{400}{100} = 4$
Using the trigonometric identity $\sec^2 \beta = 1 + \tan^2 \beta$:
$\sec^2 \beta = 1 + (4)^2 = 17$
(Note: A common student mistake is to find $\beta = \tan^{-1}(4)$ using a calculator, which introduces rounding errors and violates the 'find exactly' instruction.)

4. Substitute the known rates and values to solve for $\frac{d\beta}{dt}$.

The balloon moves horizontally away at a constant speed, so $\frac{dx}{dt} = 10\text{ m/s}$.
Substitute $x = 100$, $\frac{dx}{dt} = 10$, and $\sec^2 \beta = 17$ into the differentiated equation:
$17 \frac{d\beta}{dt} = -\frac{400}{(100)^2} (10)$
$17 \frac{d\beta}{dt} = -\frac{4000}{10000}$
$17 \frac{d\beta}{dt} = -\frac{2}{5}$
$\frac{d\beta}{dt} = -\frac{2}{85}$

1. Formulate the expression for the area of the flower bed $ABC$ in terms of $\theta$.

Let $M$ be the midpoint of $BC$. By symmetry, the altitude from $A$ to $BC$ passes through $M$ and the centre of the circle $O$.
The angle subtended by arc $BC$ at the centre is $\angle BOC = 2\angle BAC = 2\theta$.
In the right-angled $\triangle BOM$, $\angle BOM = \theta$.
$BM = OB \sin\theta = 6\sin\theta \implies BC = 2BM = 12\sin\theta$
$OM = OB \cos\theta = 6\cos\theta$
The height of $\triangle ABC$ is $AM = AO + OM = 6 + 6\cos\theta$.
The area of the flower bed, $A$, is:
$A = \frac{1}{2} \cdot BC \cdot AM$
$A = \frac{1}{2}(12\sin\theta)(6 + 6\cos\theta)$
$A = 36\sin\theta(1 + \cos\theta) = 36\sin\theta + 36\sin\theta\cos\theta$
Using the double angle identity $\sin(2\theta) = 2\sin\theta\cos\theta$:
$A = 36\sin\theta + 18\sin(2\theta)$

2. Differentiate the area function with respect to $\theta$ and set it to zero for finding stationary points.

$\frac{\mathrm{d}A}{\mathrm{d}\theta} = 36\cos\theta + 18(2)\cos(2\theta) = 36\cos\theta + 36\cos(2\theta)$
To find the maximum area, set $\frac{\mathrm{d}A}{\mathrm{d}\theta} = 0$:
$36\cos\theta + 36\cos(2\theta) = 0$
$\cos(2\theta) + \cos\theta = 0$

3. Solve the resulting trigonometric equation for acute $\theta$.

Using the double angle identity $\cos(2\theta) = 2\cos^2\theta - 1$:
$2\cos^2\theta - 1 + \cos\theta = 0$
$2\cos^2\theta + \cos\theta - 1 = 0$
$(2\cos\theta - 1)(\cos\theta + 1) = 0$
Since $\theta$ is an acute angle, $\cos\theta > 0$.
Thus, $\cos\theta = \frac{1}{2}$, which gives $\theta = \frac{\pi}{3}$.

4. Calculate the exact maximum possible area.

Substitute $\theta = \frac{\pi}{3}$ into the area expression:
$A = 36\sin\left(\frac{\pi}{3}\right) + 18\sin\left(\frac{2\pi}{3}\right)$
$A = 36\left(\frac{\sqrt{3}}{2}\right) + 18\left(\frac{\sqrt{3}}{2}\right)$
$A = 18\sqrt{3} + 9\sqrt{3} = 27\sqrt{3}$
The exact maximum area is $27\sqrt{3}\text{ m}^2$.

5. Use the second derivative test to justify that this area is a maximum.

$\frac{\mathrm{d}^2A}{\mathrm{d}\theta^2} = -36\sin\theta - 72\sin(2\theta)$
When $\theta = \frac{\pi}{3}$:
$\frac{\mathrm{d}^2A}{\mathrm{d}\theta^2} = -36\sin\left(\frac{\pi}{3}\right) - 72\sin\left(\frac{2\pi}{3}\right)$
$\frac{\mathrm{d}^2A}{\mathrm{d}\theta^2} = -36\left(\frac{\sqrt{3}}{2}\right) - 72\left(\frac{\sqrt{3}}{2}\right) = -54\sqrt{3}$
Since $\frac{\mathrm{d}^2A}{\mathrm{d}\theta^2} < 0$, the area is a maximum.