A man launches his boat from point A on a bank of a straight river, 1 km wide, a

Question

A man launches his boat from point A on a bank of a straight river, 1 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 toB, 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, where should he land to reach B as soon as possible? (We assume that the speed of the water is negligible compared to the speed at which the man rows.)

1) ______________km from C

Explanation / Answer

x = distance from C to D.

The distance from A to D = sqrt(1^2 + x^2).

The distance from D to B is 1 - x for 0 < x < 1

(It would be x - 1 for x > 1, but clearly the time increases for x > 1, as the distance is increasing for both the water and land segment. Similarly, the distance for the water and land segment are greater for x < 0 than x = 0, so this is not a solution).

Then, the time from A to D is sqrt(1^2 + x^2)/6 and

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

We seek to minimize sqrt(1^2 + x^2)/6 + (1 - x)/8

The derivative is x/2 sqrt(1^2+x^2) - 1/8

This equals 0 when x/(6 sqrt(1^2+x^2)) = 1/8

8x = 6 sqrt(1^2+x^2)

Squaring both sides, 64x^2 = 36(1 + x^2)

64x^2 = 36 + 36x^2

28x^2 = 36

x = 3/sqrt(7)

This is not in the range (0, 1). Thus, the derivative x/2 sqrt(1^2+x^2) - 1/8 is strictly negative in [0, 1] (note that it equals -1/8 at x = 0 at the endpoint)

Thus, the solution is to have D at B.

D is 1 km from C.

The time is sqrt(2)/6

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