Problem source

A ship sails at $20$ kilometres/hour in a straight line which is, at its closest, 1 kilometre from a port. A tug-boat with maximum speed 12 kilometres/hour leaves the port and intercepts the ship, leaving the port at the latest possible time for which the interception is still possible. How far does the tug-boat travel?

Solution source

The position of the ship is $\mathbf{s} = \binom{20t}{1}$. Suppose the interception is at $T$, then the ship leaves at $T-\frac1{12}\underbrace{\sqrt{400T^2+1}}_{\text{distance to intercept}}$.

We wish to maximise this, ie

\begin{align*}
&& \frac{\d}{\d T} \left ( T - \frac1{12}\sqrt{400T^2+1}\right) &= 1 - \frac{1}{12} \cdot \frac12 \cdot 400 \cdot 2T \cdot \left (400T^2+1 \right)^{-1/2} \\
&&&= 1 - \frac{100}3 T(400T^2+1)^{-1/2} \\
\Rightarrow && \frac{T}{\sqrt{400T^2+1}} &= \frac{3}{100} \\
\Rightarrow && \frac{T^2}{400T^2+1} &= \frac{9}{10000} \\
\Rightarrow && 10000T^2 &= 3600T^2+9 \\
\Rightarrow && 6400T^2 &= 9 \\
\Rightarrow && T &= \pm \frac{3}{80} \quad \text{(T > 0)}
\end{align*}

Therefore the distance is $\sqrt{400 \frac{9}{6400} + 1} = \sqrt{\frac{9}{16}+1} = \frac{5}{4} = 1.25 \text{ km}$