A man launches A man launches his boat from point A on a bank of a straight river, 3 km wide, and wants to reach point B, 1 km downstream on the opposite bank...

Question

Q. A man launches his boat from point A on a bank of a straight river, 3 km wide, and wants to reach point B, 1 km downstream on the opposite bank, as quickly as possible (see the figure below). He could row his boat directly across the river to point C and then run to B, or he could row directly to B, or he could row to some point D between C and B and then run to B. If he can row 6 km/h and run 8 km/h,how far (in km) downstream from C should he land to reach B as soon as possible?

How can I solve this problem? I want to know the answer and the process as soon as possible.

Answer 1

Picture point A at the origin, point B at (1,3), point C at (0,3). Point D will be a point x units from C towards B.

Draw a line from A to D. Can you see that this distance is represented by √(x²+9) (Pythagorean Theorem)? The distance from D to B will be represented by (1-x).

The time for each "leg" is calculated using T=D/R.

So time to get from A to D (rowing) is √(x²+9)/6, where 6 is the rowing rate (km/hr).

The time from D to B is (1-x)/8.

So the total time function looks like this:

T(x) = √(x²+9)/6 +(1-x)/8

T(0) represents the situation where the trip is row from A to C and run from C to B, i.e. C and D are the same point.

T(1) represents the situation where the trip is row directly from A to B.

The question is, is there really a point D between C and B that gives a lesser time than either of the other trips.

Find the first derivative of T(x), set it equal to zero, solve for x (critical number). If x is between 0 and 1, and that particular value of x gives a minimum time, then row to D and run from D to C. Determine the value for T(0) and T(1) and T(critical#) to find the value of x that gives the absolute minimum time.

Check it out here:

desmos.com/calculator/dttu94wcv8