A plane wreck in a desert

[neurosis]

Here's a tricky optimization problem. I'm having a hard time figuring it out and it's driving me CrAzY!! I know that I have to set the first derivative equal to zero to find the minimum; my problem is figuring out the equations involved. Anyone have any ideas? I'm getting really frustrated with it.

 

 

The wreck of a plane in a desert is 18 miles from the nearest point "A" on a straight road. A truck starts for the wreck at a point on the road that is 40 miles distant from A. If the truck can travel at 70mph on the road and at 35mph on a straight path in the desert, how far from point "A" should the truck leave the road to reach the wreck in minimum time?

[BobbyJoeCool]

v=dt, so t=v/d.

 

If you let x be what you take along the road, then you can figure what you don't take along the road... (as you have a right triangle, 18 miles, 40-x miles and the distance you shave left to travel)...

 

so your distance is [imath]x+\sqrt{(40-x)^2+18^2}[/imath].

 

And your velocity is [imath]70x+35\sqrt{(40-x)^2+18^2}[/imath].

 

so [math]t=\frac{70x+35\sqrt{(40-x)^2+18^2}}{x+\sqrt{(40-x)^2+18^2}}[/math]

 

Now... optimize!

 

EDIT: YOu get the distance by adding the distance traveled on the road, and the hypotenuse of the triangle created by taking the distance traveled on the road from 40 (which should be the distance to point A) and 18 (the distance from Point A to the fallen plane), and the velocity is just the speed multiplied by the distance traveled at each speed...