Question 909392
A ship is anchored 9 miles from the nearest point of a straight beach.
 A man in the ship wishes to row to the shore and walk to a point on the beach 17 miles from the point directly opposite the ship.
 If he can row at the rate of 4 miles per hour and walk at the rate of 5 miles per hour, where ought he to land in order to reach his destination in 4 hour and 45 minutes?
:
A right triangle is formed by the ship, point on the shore directly opposite the ship, and a point on the landing point on the shore
:
Let x = dist between the opposite point and the landing point
The rowing path will be the hypotenuse of the triangle
(17-x) = dist from the landing point to the 17 mi point
:
Write a time equation; time = dist/speed
:
Rowing time + walking time = 4.75 hrs
{{{(sqrt(x^2+9^2))/4}}} + {{{((17-x))/5}}} = 4.75
multiply by 20 to clear the denominators, results:
{{{5(sqrt(x^2+9^2))}}} + 4(17-x) = 20(4.75)
:
{{{5(sqrt(x^2+9^2))}}} + 68 - 4x = 95
:
{{{5(sqrt(x^2+81))}} = 4x + 95 - 68
:
{{{5(sqrt(x^2+81))}}} = 4x + 27
Square both sides, 
25(x^2 + 81) = 16x^2 + 216x + 729
25x^2 + 2025 = 16x^2 + 216x + 729
Combine like terms on the left
25x^2 - 16x^2 - 216x + 2025 - 729 = 0
9x^2 - 216x + 1296 = 0
Simplify, divide by 9
x^2 - 24x + 144 = 0
Factors to 
(x - 12)(x - 12) = 0
x = 12 miles from the opposite point to the landing point
:
:
Check this out, find the rowing dist and walking dist
Rowing dist = {{{sqrt(12^2+9^2)}}} = 15 mi
Walking dist = 17 - 12 = 5 mi
:
Find the times
15/4 = 3.75 hrs rowing
5/5  = 1.0 hr walking
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total time 4.75 hrs