A forest ranger is in a forest 2 miles from a straight road. A car is located c

Question

A forest ranger is in a forest 2 miles from a straight road. A car is located c miles down the road. The forest ranger can walk 3 miles per hour in the forest and travel 5 miles per hour along the road. In the picture below, the forest ranger walks in a straight line to a point x unit from the end of the road on the left, and then along the road. The total travel time for the ranger to get to the car is the sum of the travel time in the forest and the travel time on the road. Write the total time T as a function of x, using the letter c for the distance from the end of the road on the left to the car. T(x)=______ Toward what point on the road should the ranger walk in order to minimize the travel time to the car if... (a) c = 9 miles? (Numerical Answer ONLY) x= ___ (b) c = 1/5 miles? Remember that the point must lie between the left end of the highway and the car. (Numerical Answer ONLY) x=______

Explanation / Answer

The distance walked through the forest will be:
sqrt(x^2 + 4 )

The distance walked along the road will be:
c-x

Since distance equals speed*time, time = distance / speed. That allows us to get T(x).

T(x) = sqrt(x^2 + 4 )/ 3 + (c-x)/5


A)
T(x) = sqrt(x^2 + 4)/3 + (9-x)/5
T'(x) = x / (3sqrt(x^2+4)) - 1/5

Solve for x when T'(x) = 0
0 = x / (3sqrt(x^2+4)) - 1/5
x / (3sqrt(x^2+4)) = 1/5
x = 3sqrt(x^2+4)/5

x = 3/2 = 1.5 miles

B)
Repeat the same process:
T(x) = sqrt(x^2 + 4)/3 + (1/5-x)/5
T'(x) = x / (3sqrt(x^2+4)) - 1/5

Notice that if we try to solve for x when T'(x) = 0, we will again get x=1.5 and that is outside the bounds. That means that the minimum and maximum must occur on the end points. So we plug in x=0 and x=1/5 into T(x) and choose the one that is smaller.

T(0) = 2/3 + 1/25 = 53/75 .707 hours
T(1/5) = sqrt(101/25)/3 + 0/5 .670 hours

So the best choice is x = 1/5 = 0.2 miles

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