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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's denote the speed of the boat in still water as B mph.
When the boat is going upstream against the current, its effective speed is B - 6 mph.
When the boat is going downstream with the current, its effective speed is B + 6 mph.
Given that the round trip distance is 22.5 miles each way, the total distance traveled per round trip is 2 * 22.5 = 45 miles.
The time taken to travel upstream (against the current) is 22.5 / (B - 6) hours.
The time taken to travel downstream (with the current) is 22.5 / (B + 6) hours.
The total time taken for the round trip is given as 9 hours:
22.5 / (B - 6) + 22.5 / (B + 6) = 9
To solve for B, we need to solve the above equation. Let's simplify the equation:
22.5(B + 6) + 22.5(B - 6) = 9(B^2 - 36)
22.5B + 135 + 22.5B - 135 = 9B^2 - 324
45B = 9B^2 - 324
9B^2 - 45B - 324 = 0
Now we have a quadratic equation to solve. Let's divide the entire equation by 9 to simplify it:
B^2 - 5B - 36 = 0
Now let's factorize the quadratic equation:
(B - 9)(B + 4) = 0
This gives two possible values for B: B = 9 mph or B = -4 mph.
Since speed cannot be negative, the boat must travel at 9 mph in still water to meet the ferry operator's needs.