Hello, rai2003!

There's something missing from your problem.

A boat with an ill passenger is 7 1/2 mi north of a straight coastline which runs

east and west. A hospital on the coast is 60 miles from the point on shore south

of the boat. If the boat starts toward shore at 15 mph at the same time

an ambulance leaves the hospital at 60 mph and meets the ambulance

**in the least amount of time**, what is the total distance (to the nearest 0.5 mile)

traveled by the boat and the ambulance? Code:

B o
| * ___________
| * √x^2 + 7.5^2
7.5 | *
| *
| *
S o-----------------o---------------o
: - - - x - - - P - - 60-x - - H
: - - - - - - - 60 - - - - - - - :

The boat is at $\displaystyle \,B$, 7.5 miles from shore: .$\displaystyle BS = 7.5$

The hospital is at $\displaystyle \,H$: .$\displaystyle SH = 60$

The boat heads for point $\displaystyle \,P$ at 15 mph: .$\displaystyle SP = x$

Its distance $\displaystyle BP$ is the hypotenuse of a right triangle.

. . Hence: .$\displaystyle BP \,=\,\sqrt{x^2 + 7.5^2}$ miles.

At 15 mph, this will take: .$\displaystyle \dfrac{\sqrt{x^2+7.5^2}}{15}$ hours.

At the same time, the ambulance goes from $\displaystyle \,H$ to $\displaystyle \,P$ at 60 mph.

Its distance is: .$\displaystyle HP \,=\,60-x$ miles.

At 60 mph, this will take: .$\displaystyle \dfrac{60-x}{60} \:=\:1 - \dfrac{x}{60}$ hours.

The total time is: .$\displaystyle T \;=\;\dfrac{(x^2+7.5^2)^{\frac{1}{2}}}{15} + 1 - \dfrac{x}{60}$

And *that* is the function we must mininize.

Once we find $\displaystyle \,x$, we can determine the distances travelled.