A lighthouse is built on an exposed reef, 5 miles off-shore. The shoreline is perfectly straight, and a town is located 9 miles downshore from the point on the shoreline nearest the lighthouse.
The lighthouse keeper needs to go from the lighthouse to the town to get fresh supplies. He can row a boat at 1.8 miles per hour, and run at 3.6 miles per hour. How far downshore towards the town should he land, if he wants to get to the town as quickly as possible?
Distance downshore to the landing point =
sqrt(25/3) miles.
Time it takes to get from the lighthouse to the town = 4.906 hours.
After several such trips, the lighthouse keeper buys a motor boat, which can go 6 miles per hour.
How far downshore should he land if he uses his motorboat, instead of rowing?
Just substitute 6 for 1.8 in your time function. You will find that the shortest time is to boat all the way.
A little analysis will show you that if the boat is faster than the running, the boat is the best way. If the ratio of the boat:run speed is k, then the distance downshore for fastest time is x=5k/√(1-k^2)
In the original problem, k = 1/2, so x = 5/√3 as you already found.
To determine how far downshore the lighthouse keeper should land if he uses his motorboat, we need to find the optimal landing point that minimizes the total time it takes for him to reach the town.
Let's calculate the time it would take for the lighthouse keeper to row from the lighthouse to the landing point, and then run from the landing point to the town.
Time to row from the lighthouse to the landing point:
Distance = 5 miles
Speed = 1.8 mph
Time = Distance / Speed = 5 / 1.8 = 2.7778 hours
Time to run from the landing point to the town:
Distance = 9 - x miles (where x is the distance downshore to the landing point)
Speed = 3.6 mph
Time = Distance / Speed = (9 - x) / 3.6
So, the total time it takes for the lighthouse keeper to reach the town using the rowboat is:
Total Time = Time to row + Time to run = 2.7778 + (9 - x) / 3.6
Now let's consider the case where the lighthouse keeper uses his motorboat.
Time to motorboat from the lighthouse to the landing point:
Distance = 5 miles
Speed = 6 mph
Time = Distance / Speed = 5 / 6 = 0.8333 hours
Time to run from the landing point to the town:
Distance = 9 - x miles (same as before)
Speed = 3.6 mph (same as before)
Time = Distance / Speed = (9 - x) / 3.6
So, the total time it takes for the lighthouse keeper to reach the town using the motorboat is:
Total Time = Time to motorboat + Time to run = 0.8333 + (9 - x) / 3.6
To find the optimal landing point, we need to minimize the total time. So, we need to find the value of x that minimizes the expression for the total time:
Total Time = 2.7778 + (9 - x) / 3.6
To find the minimum, we can take the derivative of this expression with respect to x and set it to zero:
d(Total Time) / dx = -(1/3.6) = 0
Simplifying this equation, we get:
1/3.6 = 0
Since this is not possible, it means that the total time is already at its minimum when x = 0. Therefore, the lighthouse keeper should land directly at the point on the shoreline nearest the town.
In other words, if he uses his motorboat, he should land right at the town, which is 9 miles downshore from the lighthouse.