Optimization question...

A ship is sailing due north at 12km/h while another ship is observed 15km ahead, travelling due east at 9 km/h. What is the closest distance of approach of the two ships?

i know that we are trying to minimize the distance between the two ships...and that the shortest distance between them will the the perpendicular distance...but after that i'm lost

Like what formula's do I use?

------------>9km/hr ((b= 9t)
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^
12km/hr
a= 15-12t

The path forms a right triangle. The distance between the ships is the hypotenuse.
So a^2+b^2=c^2
solve for c in terms of t. Minimize c by setting c' = 0

C= sqr[(9t)^2+(15-12t)^2]
= sqr[225t^2-360t+225]
c'= {(1/2)[225t^2-360t+225]^(-1/2)}*
(450t-360)= 0.... I dont know why yahoo cuts off the equation?

For a fraction to equal zero, set the numerator = 0
450t-360=0
t=360/450=4/5 hr
Now plug in t to find c
I got 9 km
Hope this helps!

It's not too bad....

You have to figure out how long it will take the first ship to travel 15km. This is done by using the old distance/rate = time formula.

So 15km / 12km/h = 1.25 h. In 1.25 h, the first ship will cross the path (behind) of the east-bound ship. At this point, the eastbond-ship will be due east, but by how much???

Now, in 1.25 hours, the east-bound ship will be 9km/h * 1.25h = 11.25 km away from it's point that was ahead of the first ship.

So your answer is 11.25 km.