Question

An island is at point A 6 mi offshore from the nearest point B on a straight beach. A store is at point C 7 mi down the beach from B. If a man can row at the rate of 4 mi/hr, and walk at the rate of 5 mi/hr, where should he land in order to go from the island to the store in the least possible time.

Answer 1

Let D be a point between B and C and let the distance BD = x, we note that speed = d / t which implies that t = d/speed where d = distance and t = time

By Pythagoras theorem

$AD^2=x^2+6^2 \implies AD=\sqrt{x^2+6^2}\\Also \text{ DC = 7-x}\\\text{Therefore Time taken from A to D }=\frac{\sqrt{x^2+6^2}}{4}\\\text{Therefore Time taken from D to C }=\frac{7-x}{5}\\\text{Therefore total time taken }= \frac{\sqrt{x^2+6^2}}{4} + \frac{7-x}{5}\\\text{Differentiating equation above we have that }\frac{dt}{dx} = \frac{x}{4\sqrt{x^2+6^2}}-\frac{1}{5}\\\text{Set } \frac{dt}{dx}=0, \implies \frac{5x-4\sqrt{x^2+6^2}}{20\sqrt{x^2+6^2}}=0\\\implies 5x-4\sqrt{x^2+6^2} =0\\\text{therefore } 5x=4\sqrt{x^2+6^2} \text{ squaring both sides we have that } 25x^2 = 16x^2 +16*36\\x=8mi\\\text{AD}=\sqrt{8^2+6^2}\implies AD=10mi\\\text{Hence the man would land the boat 10mi from A to get to the store in the least possible time}$