Question

A man is in a boat 2 miles from the nearest point on the coast. He is to go to point Q, located 3 miles down the coast and 1 mile inland (see figure). He can row at a rate of 4 miles per hour and walk at 4 miles per hour. Toward what point on the coast should he row in order to reach point Q in the least time?

Answer 1

Answer:

The answer is "0.45385".

Step-by-step explanation:

The time is taken to reach the coast $= \frac{\sqrt{(2^2 + x^2)}}{4}$

The  time is taken to reach Q after reaching the coast $= \frac{\sqrt{(1 + (3-x)^2)}}{4}$

Total time, $T= \frac{\sqrt{(1 + (3-x)^2)}}{4}+ \frac{\sqrt{(1 + (3-x)^2)}}{4}$

This has to be minimum,

$\to \frac{dt}{dx} = 0$

$\to \frac{\sqrt{(2^2 + x^2)}}{2} + \frac{\sqrt{1 + (3-x)^2}}{3}\\\\ \frac{x}{4}\sqrt{(4+x^2)} + \frac{(x-3)}{(4\sqrt{(x^2-6x+10)}} = 0\\\\\frac{x^2}{16} \times (4+x^2) = \frac{(x-3)^2}{16 \times (x^2-6x+10)}\\\\x^2(4+x^2) = \frac{(x-3)^2}{(x^2-6x+10)}\\\\4x^2+x^4 = \frac{ x^2+9-6x}{(x^2-6x+10)}\\\\ 4x^2+x^4(x^2-6x+10) = x^2+9-6x\\\\4x^4-24x^3+40x^2+x^6-6x^5+10x^4= x^2+9-6x\\\\x^6-6x^5+ 14x^4-24x^3+39x^2+6x-9=0$

x=0.45385