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?..
Let the speed of the boat in still water be x mph.
When the boat is traveling against the current, its effective speed is (x - 6) mph.
And when the boat is traveling with the current, its effective speed is (x + 6) mph.
Since the total distance covered by the boat is 22.5 miles each way, the total distance covered is 45 miles.
The time taken to travel 22.5 miles against the current is 22.5 / (x - 6) hours.
Similarly, the time taken to travel 22.5 miles with the current is 22.5 / (x + 6) hours.
The total time taken for the round trip is given as 9 hours.
Therefore, 22.5 / (x - 6) + 22.5 / (x + 6) = 9
Multiplying through by (x - 6)(x + 6) to clear the fractions, we get:
22.5(x + 6) + 22.5(x - 6) = 9(x^2 - 36)
45x = 9x^2 - 324
Rearranging, we get:
9x^2 - 45x - 324 = 0
Dividing throughout by 9:
x^2 - 5x - 36 = 0
Solving this quadratic equation, we get:
(x - 9)(x + 4) = 0
This gives us two possible values of x: x = 9 mph and x = -4 mph.
Since the speed of the boat cannot be negative, the speed of the boat in still water must be 9 mph.
Therefore, the boat must go at a speed of 9 mph on the lake in order for it to serve the ferry operator's needs.