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Expert: - 4/19/2010


Question
Cowboy Curtis wants to build a dirt road from his ranch to the highway so that he can drive to the city in the shortest amount of time. The perpendicular distance from the ranch to the highway is 4 miles and the city is located 9 miles down the highway. Where should Curtis join the dirt road to the highway if the speed limit is 20 mph on the dirt road and 55 mph on the highway?

Answer
Take the road as entering the highway 9-x units from town.

This means that the length of the road on his property is the hypoteneuse of a triangle wit edges of length x and length 4.  Thus, the road on his property would be √(x²+16).

Since it is known that d = rt, where d is distance, r is rate, and t is time,
we know that on the highway we have 9-x = 55*(t1) and on the farm it is √(x²+16) = 10*(t2).
The goal of this problem is to minimize the time, which it t1+t2.

The equations tell us that t1 = (9-x)/55 and t2 = [√(x²+16)]/10.
Total time is the sum of the two, so t(x) = (9-x)/55 + [√(x²+16)]/10.

Take the derivative, set it to 0, and solve for x.

The value of x will say how far down the road from straight out to the highway
he should make the road on his property hit the highway.  I believe the best way
to do this is to go to town and measure 9-x miles out.

Calculus

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