Question 40450
b = rate of boat
r = rate of river
There are 2 combined distances that the boat travels. Ther is the 1 mile
distance and the distance beyond 1 mile which takes the boat 5 minutes
going against the current.
{{{d(1) = 1}}}
{{{d(2) = (b - r)*5}}}
b - r is the boats rate going upstream
(b - r)*5 is rate x time  = distance
{{{d(1) + d(2)}}} is the combined distance
When the boater goes back to get the hat, he will go back to his starting 
point at a rate of (b + r) going with the current. His time is the 
combined distances divided by his rate.
{{{t = (1 + (b - r)*5) / (b + r)}}}
During this time, the hat floats downstream for 1 mile back to the 
starting point at the rate the river is flowing.
{{{t = 1 / r}}}
These times are equal so,
{{{t = (1 + (b - r)*5) / (b + r) = 1 / r}}}
{{{r + r(b - r)*5 = b + r}}}
{{{r + (br - r^2)*5 = b + r}}}
{{{-5r^2 + 5br - b = 0}}}
{{{5r^2 -5br +b = 0}}}
{{{r^2 - br + b/5 = 0}}}
complete the square
{{{r^2 - br + (b^2)/4 = (b^2)/4 - b/5}}}
{{{(r - b/2)^2 = (b^2)/4 - b/5}}}
{{{r - b/2 = 0+-sqrt(((b^2)/4) - b/5)}}}
{{{r = b/2 +-sqrt(((b^2)/4) - b/5)}}}
Try different values for b, speed of boat to get values for r