Math: Optimization Problem?

Question

Hey everyone! I really need some help here. I've been working on it all this week now and I can't seem to get an answer. Any help would be much appreciated, thank you.

****I have added a link to the sketch of the problem for the visualization here:

http://i11.photobucket.com/albums/a197/arminvanbuuren/feedlotproblem.gif

There's a feedlot 200 feet long with a water trough along one edge and a feed bin located on an adjacent angle. A cow enters the gate at point A (90 vertical feet from the water trough). She walks straight to point P, gets a drink from the trough, and then walks straight to the feed bin at point B (60 vertical feed from the water trough). If the cow knew calculus, what point P along the water trough would she select to minizmize the total distance she walks?

Answer 1

let x be the horizontal distance from the gate side to P
so that 200-x is the horizontal distance from P to the bin side
then AP = sqrt(x^2 + 90^2) by pythagoras
and PB = sqrt((200-x)^2 + 90^2)
adding gives
D= sqrt(x^2 + 90^2) + sqrt((200-x)^2 + 90^2)
0=dD/dx=2x/ sqrt(x^2 + 90^2) - 2(200-x)/sqrt((200-x)^2 + 90^2)
then take the 2(200-x)/sqrt((200-x)^2 + 90^2) term to the otherside, square both sides and solve the horrible equation for x which I won't do cause it will take me hours
have fun!
lol!

Answer 2

Let P be x feet from the side of the feedlot with the gate in it. Then the distance AP is √(90^2 + x^2) feet and the distance PB is √(60^2 + (200-x)^2) feet.
So the total distance travelled is D(x) = √(90^2 + x^2) + √(60^2 + (200-x)^2).
D'(x) = (1/2)(90^2 + x^2)^(-1/2) (2x) + (1/2) (60^2 + (200-x)^2)^(-1/2) (2(200-x)(-1))
= x / √(90^2 + x^2) - (200-x) / √(60^2 + (200-x)^2)
= 0 <=> x √(60^2 + (200-x)^2) = (200-x) √(90^2 + x^2)
Square both sides to get
x^2 (60^2 + (200-x)^2) = (200-x)^2 (90^2 + x^2)
<=> 3600x^2 + x^2 (200-x)^2 = 8100(200-x)^2 + x^2 (200-x)^2
<=> 3600x^2 = 8100(40000 - 400x + x^2)
<=> 4x^2 = 9(40000 - 400x + x^2)
<=> 5x^2 - 3600x + 360000 = 0
<=> x^2 - 720x + 72000 = 0
<=> (x - 120) (x - 600) = 0
We know x will be between 0 and 200, so we take x = 120.

So the distance is minimised if P is 120 feet along the water trough.