Where should he land if he can row 2 mi/h and walk 4 mi/h?

Question

A man in a rowboat at P is 5 mi from the nearest point A on a straight shore, wishes to reach a point B, 6 mi from A along the shore, in the shortest time. Where should he land if he can row 2 mi/h and walk 4 mi/h?

Answer 1

Hi Geroy,

I hope this is helpful -- this is a classic calculus problem that is really hard to explain without a drawing, but I will try my best!

Let's define some terms first.

Let's pick a point between points A and B where the man lands the boat and call it point C.

Now, let's call the distance between A and C, x. Then we know that the distance the man will walk (the distance between B and C, "BC") becomes 6 - x.

Notice that the triangle APC is a right triangle. (∠PAC is the right angle)

Thus, we can find out the distance that the man will row. (the distance between P and C, "PC")

Use the Pythagorean theorem, and you will get PC = √(25+x²).

We know that (distance) = (time) * (rate), so to find time we need (time) = (distance) / (rate).

Let's call the time it takes the man to row, t_r and to walk, t_w.

Then t_r = √(25+x²) / 2 and t_w = (6 - x) / 4. Thus, t_total = √(25+x²) / 2 + (6 - x) / 4.

We want to minimize t_total. To do so, we need to take the derivative of t_total with respect to x.

[ t_total] d / dx = x / 2√(25 + x²) - 1 / 4 and set the derivative to 0. Then you get x = √(25/3) ≈ 2.89.

Therefore he should land 2.89 miles away from point A towards point B.

Hope this was helpful.

Best,

Sunghyun

Answer 2

Draw and label a diagram.

L represents the landing point

AL = x

LB = 6-x

PL = √(25 + x²)

f(t) = [(6 - x) / 4] + [√(25 + x²) / 2]

Determine f'(t) and set equal to 0.