A hunter is at a point on a river bank. he wants to get his cabin, located 19 miles north and 8 miles west. He can travel 5 mph on the river bank and 2 mph on the rough rocky ground.how far upriver should he go in order to reach the cabin in the minimum amount of the time.

To find out how far upriver the hunter should go in order to reach the cabin in the minimum amount of time, we need to determine the most efficient route.

First, let's calculate the time it would take the hunter to reach the cabin by traveling directly towards it.

The distance he needs to travel north is 19 miles, and his speed on the river bank is 5 mph. So it would take him 19 miles / 5 mph = 3.8 hours to travel north.

Now, let's calculate the time it would take the hunter to travel upriver.

Since the hunter can only travel 2 mph on rough rocky ground, he will need to go upriver until the ratio of time spent on the river bank to time spent on rough rocky ground is equal to 5/2 (his speeds on the river bank and rough rocky ground).

Let's assume the distance he travels upriver is X miles. The remaining distance he would need to travel west from that point is (8 - X) miles.

The time it would take the hunter to travel north and west using this route can be calculated as follows:

Time on the river bank = distance on the river bank / speed on the river bank = X miles / 5 mph = X/5 hours

Time on rough rocky ground = distance on rough rocky ground / speed on rough rocky ground = (8 - X) miles / 2 mph = (8 - X)/2 hours

To minimize the total time, we need to find the value of X that makes the sum of the two times (X/5 + (8 - X)/2) as small as possible.

Let's simplify the equation:

X/5 + (8 - X)/2 = (2X + 5(8 - X)) / 10
= (2X + 40 - 5X) / 10
= (40 - 3X) / 10

To find the minimum value, we can use calculus by taking the derivative of the equation with respect to X and setting it equal to zero:

d/dX [(40 - 3X) / 10] = -3/10

Setting -3/10 equal to zero, we find that X = 40/3 miles.

Therefore, the hunter should go 40/3 miles upriver to reach the cabin in the minimum amount of time.