下图显示了仓库 $ABCD$ 的侧视图，其中 $BC = 10\text{ m}$，$CD = 6\text{ m}$，$\angle BCD = \angle BAD = 90^\circ$ 且 $\angle CBA = \theta$。

1. Set up the R-formula equation

Let $6 \sin \theta + 10 \cos \theta = R \sin(\theta + \alpha) = R(\sin \theta \cos \alpha + \cos \theta \sin \alpha)$.
Comparing coefficients:
$R \cos \alpha = 6$
$R \sin \alpha = 10$

2. Find the value of R

$R = \sqrt{6^2 + 10^2} = \sqrt{36 + 100} = \sqrt{136} = 2\sqrt{34}$

3. Find the value of alpha

$\tan \alpha = \frac{10}{6} = \frac{5}{3}$
$\alpha = \tan^{-1}\left(\frac{5}{3}\right) \approx 59.036^\circ$
So, $AB = 2\sqrt{34} \sin(\theta + 59.0^\circ)$

4. Find the maximum value of AB

The maximum value of $\sin(\theta + 59.036^\circ)$ is $1$.
Maximum value of $AB = 2\sqrt{34} \approx 11.7$ m

5. Find the value of theta for which the maximum occurs

$\sin(\theta + 59.036^\circ) = 1$
$\theta + 59.036^\circ = 90^\circ$
$\theta = 90^\circ - 59.036^\circ = 30.964^\circ \approx 31.0^\circ$

1. Identify the condition for the maximum value of AB

For $AB$ to be maximum, $\sin(\theta + \alpha) = 1$.

2. Explain why the obtained angle is the smallest possible

The general solution for $\sin(\theta + \alpha) = 1$ is $\theta + \alpha = 90^\circ + 360^\circ k$ for integer $k$. The smallest positive value of $\theta + \alpha$ is $90^\circ$ (when $k=0$), which consequently gives the smallest positive value for $\theta$.