To solve this problem, we must find the optimal path for the person to intercept the boat in the minimum possible time. The maximum speed of the boat, $v_{max}$, is such that the time it takes for the boat to reach the interception point P is equal to this minimum time for the person.

1. Optimal Path for the Person Let the person run from A to B and then swim from B to P, intercepting the boat at point P. The total time for the person is $t_{person} = t_{AB} + t_{BP} = \frac{AB}{v_1} + \frac{BP}{v_2}$. For a given interception point P, the person must choose point B on the shore to minimize this time. This is analogous to Fermat's principle for light refraction. The condition for the minimum time is that the angle of swimming, $\theta$, with respect to the shore satisfies:

$$ \cos \theta = \frac{v_2}{v_1} $$

For a path to be optimal, the time saved by running an infinitesimal extra distance $dx$ along the shore must be balanced by the extra time taken to swim. This leads to the condition above, which holds for any viable interception. If $v_2 > v_1$, the person should swim directly from A. Here, $v_2 < v_1$, so this mixed path is possible.

2. Kinematic Condition for Interception Let the interception occur at time $t$. The distance traveled by the boat is $AP = vt$. The time taken by the person following the optimal path is $t_{person, min}$. For the limiting case of maximum boat speed, $v_{max}$, we have:

$$t = \frac{AP}{v_{max}} = t_{person, min} = \frac{AB}{v_1} + \frac{BP}{v_2}$$

3. Geometric Analysis Consider the triangle ABP. The angles are:

• $\angle PAB = \alpha$
• $\angle ABP = 180^\circ - \theta$
• $\angle APB = 180^\circ - \alpha - (180^\circ - \theta) = \theta - \alpha$

Using the Law of Sines on triangle ABP:

$$ \frac{AB}{\sin(\theta - \alpha)} = \frac{BP}{\sin \alpha} = \frac{AP}{\sin(180^\circ - \theta)} = \frac{AP}{\sin \theta} $$

From this, we express the path lengths $AB$ and $BP$ in terms of $AP$:

$$ AB = AP \frac{\sin(\theta - \alpha)}{\sin \theta} $$ $$ BP = AP \frac{\sin \alpha}{\sin \theta} $$

4. Derivation of Maximum Speed Substitute these expressions into the time equality:

$$ \frac{AP}{v_{max}} = \frac{1}{v_1} \left( AP \frac{\sin(\theta - \alpha)}{\sin \theta} \right) + \frac{1}{v_2} \left( AP \frac{\sin \alpha}{\sin \theta} \right) $$

Canceling $AP$ from both sides:

$$ \frac{1}{v_{max}} = \frac{\sin(\theta - \alpha)}{v_1 \sin \theta} + \frac{\sin \alpha}{v_2 \sin \theta} $$

Using the angle subtraction formula $\sin(\theta - \alpha) = \sin\theta \cos\alpha - \cos\theta \sin\alpha$:

$$ \frac{1}{v_{max}} = \frac{\sin\theta \cos\alpha - \cos\theta \sin\alpha}{v_1 \sin \theta} + \frac{\sin \alpha}{v_2 \sin \theta} $$ $$ \frac{1}{v_{max}} = \frac{\cos \alpha}{v_1} - \frac{\cos \theta \sin \alpha}{v_1 \sin \theta} + \frac{\sin \alpha}{v_2 \sin \theta} = \frac{\cos \alpha}{v_1} + \sin \alpha \left( \frac{1}{v_2 \sin \theta} - \frac{\cos \theta}{v_1 \sin \theta} \right) $$

Now, we use the optimal path condition, $\cos \theta = v_2 / v_1$. From this, we find $\sin \theta$:

$$ \sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - \left(\frac{v_2}{v_1}\right)^2} = \frac{\sqrt{v_1^2 - v_2^2}}{v_1} $$

Substitute $\cos \theta$ and $\sin \theta$ into the equation for $1/v_{max}$:

$$ \frac{1}{v_{max}} = \frac{\cos \alpha}{v_1} + \sin \alpha \left( \frac{1}{v_2 \frac{\sqrt{v_1^2 - v_2^2}}{v_1}} - \frac{v_2/v_1}{v_1 \frac{\sqrt{v_1^2 - v_2^2}}{v_1}} \right) $$ $$ \frac{1}{v_{max}} = \frac{\cos \alpha}{v_1} + \frac{\sin \alpha}{\frac{\sqrt{v_1^2 - v_2^2}}{v_1}} \left( \frac{1}{v_2} - \frac{v_2}{v_1^2} \right) $$ $$ \frac{1}{v_{max}} = \frac{\cos \alpha}{v_1} + \frac{v_1 \sin \alpha}{\sqrt{v_1^2 - v_2^2}} \left( \frac{v_1^2 - v_2^2}{v_1^2 v_2} \right) $$ $$ \frac{1}{v_{max}} = \frac{\cos \alpha}{v_1} + \frac{\sin \alpha \sqrt{v_1^2 - v_2^2}}{v_1 v_2} $$ $$ \frac{1}{v_{max}} = \frac{v_2 \cos \alpha + \sqrt{v_1^2 - v_2^2} \sin \alpha}{v_1 v_2} $$

The maximum speed of the boat is:

$$ v_{max} = \frac{v_1 v_2}{v_2 \cos \alpha + \sqrt{v_1^2 - v_2^2} \sin \alpha} $$

5. Numerical Calculation Given values: $v_1 = 4$ m/s, $v_2 = 2$ m/s, $\alpha = 15^\circ$.

$$ \sqrt{v_1^2 - v_2^2} = \sqrt{4^2 - 2^2} = \sqrt{12} = 2\sqrt{3} \, \text{m/s} $$

The trigonometric values for $15^\circ$ are:

$$ \cos 15^\circ = \frac{\sqrt{6} + \sqrt{2}}{4}, \quad \sin 15^\circ = \frac{\sqrt{6} - \sqrt{2}}{4} $$

Substitute these into the expression for $v_{max}$:

$$ v_{max} = \frac{(4)(2)}{2 \left(\frac{\sqrt{6} + \sqrt{2}}{4}\right) + 2\sqrt{3} \left(\frac{\sqrt{6} - \sqrt{2}}{4}\right)} $$ $$ v_{max} = \frac{8}{\frac{\sqrt{6} + \sqrt{2}}{2} + \frac{\sqrt{18} - \sqrt{6}}{2}} = \frac{16}{\sqrt{6} + \sqrt{2} + 3\sqrt{2} - \sqrt{6}} $$ $$ v_{max} = \frac{16}{4\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2} \, \text{m/s} $$