Question

1.A woman is on a lake in a canoe 1km from the closest point P of a straight shoreline. She wants to get to a point Q, 10km along the shore from P. To do so, she paddles to a point R between P and Q and then walks the remaining distance to Q. She can paddle 3km/hour and walk 5km/hour. How should she pick the point R so that she gets to Q as quickly as possible?

Answer 1

Question #6736 .A woman is on a lake in a canoe 1km from the closest point P of a straight shoreline. She wants to get to a point Q, 10km along the shore from P. To do so, she paddles to a point R between P and Q and then walks the remaining distance to Q. She can paddle 3km/hour and walk 5km/hour. How should she pick the point R so that she gets to Q as quickly as possible?

Solution. Suppose that now she is at the point $X$. then $PX=1$, suppose that the desirable point $R\in[P,Q]$ and $RP=x$. Due to Pythagorean theorem $XR=\sqrt{1+x^{2}}$. The time womand spends on paddling from $X$ to $R$ and walking from $R$ to $Q$(that is $10-x$ km):

$f(x):=\frac{\sqrt{1+x^{2}}}{3}+\frac{10-x}{5}$

Minimizing the last with respect to $x$. Take derivative $f^{\prime}(x)=\frac{x}{3\sqrt{1+x^{2}}}-1/5=0$, hence $x=3/4$. On this plan she will spend 34/15 h. We also must chek points $x=0$ and $x=10$(this obviously does not interest us, as t will certainly give bigger time). $f(0)=1/3+10/5=35/15>34/15$.

Answer $PR=3/4$.

##