The speed of the current in a river is 6 mph. A ferry operator who works that part of the river is looking to buy a new boat for his business. Every day, his route takes him 22.5 miles each way against the current and back to his dock, and he needs to make this trip in a total of 9 hours. He has a boat in mind, but he can only test it on a lake where there is no current. How fast must the boat go on the lake in order for it to serve the ferry operator's needs?
To determine how fast the boat must go on the lake in order to meet the ferry operator's needs, we first need to calculate the speed the boat needs to travel relative to the water.
Let's denote the speed of the boat in still water as B mph. Against the current, the speed of the boat will be B - 6 mph (since the current is flowing against the boat's direction). With the current, the speed of the boat will be B + 6 mph.
The ferry operator travels 22.5 miles each way against the current and back, so the total distance of the round trip is 45 miles. To make this trip in 9 hours, we can set up the following equation:
22.5 / (B - 6) + 22.5 / (B + 6) = 9
Multiplying both sides by (B - 6)(B + 6) to get rid of the denominators, we have:
22.5(B + 6) + 22.5(B - 6) = 9(B - 6)(B + 6)
Expanding the equation:
22.5B + 135 + 22.5B - 135 = 9(B^2 - 36)
45B = 9B^2 - 324
Rearranging the equation to form a quadratic equation:
9B^2 - 45B - 324 = 0
Divide by 9 to simplify:
B^2 - 5B - 36 = 0
Factor the quadratic equation:
(B - 9)(B + 4) = 0
B = 9 mph or B = -4 mph
Since the speed of the boat cannot be negative, the speed of the boat in still water must be 9 mph in order to meet the ferry operator's needs.