# Optimizing travel time

A woman is on a boat 4 miles north of shore. There is a restaurant 6 miles east of her current position. Her boat can move at 3 miles per hour, and she can walk at 2 miles per hour. Find the distance from the restaurant that she will land on shore so that her time is minimized.

My attempt: (Incorrect, the answer is $\frac{8}{\sqrt{5}}$ miles)

I drew a right angle triangle with horizontal component $x$ and vertical component $4$.

The distance she then has to travel from shore to the restaurant is $6 - x$

The distance she travels from the water to the shore is then (right angle triangle) $\sqrt{4^2 + x^2}$

So, I figured the time she has to travel is given by $t(x)=\frac{\sqrt{4^2 + x^2}}{3} + \frac{6-x}{2} \implies t'(x)= \frac{2x-3 \sqrt{x^2+16}}{6 \sqrt(x^2+16)}$

There are no real solutions for $t'(x)=0$ or $t'(x)$ D.N.E. I am unable to solve this problem and have been stuck on it for several hours.

EDIT: I also checked both endpoints, with $x=0$ or $x=6$ and I still don't get the right answer of $x=\frac{8}{\sqrt{5}}$

• There's no minimum because the boat is both faster and can go straight (no trade-off needed, it's just better). So just take the entire path by boat, therefore $x=6$. Oct 30 '14 at 23:01
• There is likely a typo in the question. Intuitively, the shortest path between two points is a straight line. Given that she travels faster by boat then by foot, there is no incentive for her to walk at all, yielding $x = 6$. Oct 30 '14 at 23:01

You are doing fine. Your range of $x$ is restricted to $[0,6]$ as if she lands farther than the restaurant the walking time becomes $\dfrac {x-6}3$ The fact that there is no zero of the derivative says the minimum comes at one end of the interval, here at $x=6$ as shown by this Alpha plot. She should travel on the water directly to the restaurant, so the distance is zero. I don't get $\dfrac{8}{\sqrt 5}$ even if I reverse the speeds.