Question 486052
: A boardwalk is parallel to and 210 ft inland from a straight shoreline.
 A sandy beach lies between the boardwalk and the shoreline.
 A man is standing on the boardwalk, exactly 750 ft across the sand from his beach umbrella, which is right at the shoreline.
 The man walks 4ft/s on the boardwalk and 2 ft/s on the sand.
How far should he walk on the boardwalk before veering off onto the sand if he wishes to reach his umbrella in exactly 4 min 45 s?
:
Change 4 min 45 sec to 285 sec
:
Find the distance from the man to a point (p) on the boardwalk directly opposite the umbrella.
:
p = {{{sqrt(720^2 - 210^2)}}}
p = 720 ft 
;
Let x = the dist the man will walk on the boardwalk, before onto the sand
The distance walked on the sand will be the hypotenuse of a triangle with the legs of 210' and (720-x) or {{{sqrt(210^2 + (720-x)^2)}}}
:
Write the time equation
b.w. time + sand walk time = 285 sec
{{{x/4}}} + {{{(sqrt(210^2 + (720-x)^2))/2}}} = 285
:
Get rid of the denominators, mult by 4, results
x + {{{2*sqrt(210^2 + (720-x)^2)}}} = 1140
:
After a lot math obtained a quadratic equation of:
3s^2 - 3480x + 950400 = 0
Two solutions
x = 720
x = 440 ft. the reasonable solution is the distance he walked on the boardwalk