A man launches his boat from point A on a bank of a straight river, 3 km wide, and wants to reach point B, 2 km downstream on the opposite bank, as quickly as possible. 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 B and C 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.)

Well, it sounds like this man is on a mission to reach point B as quickly as possible! Let's crunch some numbers and figure out his best approach.

If he chooses to row directly across the river to point C, he would have to row a total distance of 3 km. So it would take him 3 km / 6 km/h = 0.5 hours to row to point C. Then, he would have to run 2 km / 8 km/h = 0.25 hours from point C to point B. Therefore, the total time for this approach would be 0.5 hours + 0.25 hours = 0.75 hours.

If he decides to row directly to point B, he would cover a straight line distance of 2 km. Rowing at a speed of 6 km/h, it would take him 2 km / 6 km/h = 1/3 hours to reach point B.

Now, let's consider the third option: rowing to some point D between B and C. This is a bit of a wild card as we don't yet know which point D would be the fastest. Let's assume he rows to a point D that is exactly halfway between B and C. In this case, he would row a distance of 1.5 km. It would take him 1.5 km / 6 km/h = 1/4 hours to reach point D. Then, he would need to run a distance of 0.5 km / 8 km/h = 1/16 hours from point D to point B. Therefore, the total time for this approach would be 1/4 hours + 1/16 hours = 5/16 hours.

Comparing the three options, we can see that rowing directly to point B would take the least amount of time at 1/3 hours, while rowing to point D would take the most time at 5/16 hours, and rowing to point C would be in the middle at 0.75 hours.

So, to reach point B as soon as possible, the man should row directly to point B. It seems like taking the shortest path is the quickest way to his destination in this case!

To determine the best landing point for the man to reach point B as quickly as possible, we need to compare the times it would take for each option.

Option 1: Rowing directly across the river to point C and then running to B.
In this case, the man would need to row a straight line from point A to point C, which is 3 km wide. Since the man rows at a speed of 6 km/h, it would take him 3 km / 6 km/h = 0.5 hours to reach point C. He would then need to run from point C to point B, which is 2 km downstream. Since he runs at a speed of 8 km/h, it would take him 2 km / 8 km/h = 0.25 hours to reach point B. Therefore, the total time for option 1 is 0.5 hours + 0.25 hours = 0.75 hours.

Option 2: Rowing directly to point B.
In this case, the man would need to row a straight line from point A to point B, which is 2 km downstream. Since the man rows at a speed of 6 km/h, it would take him 2 km / 6 km/h = 0.33 hours to reach point B.

Option 3: Rowing to some point D between B and C and then running to B.
In this case, the man would need to row a straight line from point A to point D, which is a distance x downstream from point A. The width of the river at point D would be (3 - x) km. The time it would take to row to point D would be x km / 6 km/h = x/6 hours. After reaching point D, the man would need to run a distance of (3 - x) km to reach point B. This would take (3 - x) km / 8 km/h = (3 - x)/8 hours. Therefore, the total time for option 3 is (x/6) hours + ((3 - x)/8) hours.

To find the optimal landing point D, we equate the times for options 2 and 3 and solve for x:

0.33 hours = (x/6) hours + ((3 - x)/8) hours

Multiplying through by the LCD (24) gives:

7.92 = 4x + 3 - x

Simplifying the equation gives:

7.92 = 3x + 3

Subtracting 3 from both sides gives:

4.92 = 3x

Dividing by 3 gives:

x = 1.64

Therefore, the optimal landing point D would be 1.64 km downstream from point A. The man should row to this point and then run the remaining distance to point B. This would result in the minimum total time and allow him to reach point B as quickly as possible.

I assume you made a diagram according to your description.

Let CD = x km, 0 ≤ x ≤ 2
then DB = 2-x km
let AD = y
then y= (x^2 + 9)^(1/2)

time rowing = (x^2 + 9)^(1/2)/6
time running = (2-x)/8

Total Time = (x^2 + 9)^(1/2)/6 + (2-x)/8
d(Total Time)/dx = x/[6((x^2 + 9)^(1/2)] - 1/8 = 0 for a max/min of TT

This simplified to
8x = 6√(x^2+9)
4x = 3√(x^2+9) , I then squared both sides
16x^2 = 9x^2 + 81
x = 3.4

but that is outside of our domain,
unless I made an arithmetic error.
Better check it.

So let's check our trivial routes:
all rowing: y = √13
time = √13/6 = .6 hrs
combination rowing + running
time = 3/6 + 2/8 = .75 hours

So he should just row directly to B

Draw the diagram. Label the points. Now label DB as x, and CD as 2-x

So the distance AD is sqrt(3^2+(2-x)^2)

so the time for the trip is

time=DB/6 + x/8
= 1/6 (sqrt(above )) + x/8
take the deriviative of time with respect to x, set to zero, and solve for x.