Question

Your task is to build a road joining a ranch to a highway that enables drivers to reach the city in the shortest time. The perpendicular distance from the ranch to the highway is 30 km and the city is 50 km due east down the highway.
(a) Draw a picture that describes this situation and would help you determine how this should be done if the speed limit is 60 km/h on the road and 110 km/h on the highway.

Answer 1

Answer:

(a)In the attachment

(b)The road of length 35.79 km should be built such that it joins the highway at 19.52km from the perpendicular point P.

Step-by-step explanation:

(a)In the attachment

(b)The distance that enables the driver to reach the city in the shortest time is denoted by the Straight Line RM (from the Ranch to Point M)

First, let us determine length of line RM.

Using Pythagoras theorem

$|RM|^{2}=30^2+x^2\\|RM|=\sqrt{30^2+x^2}$

The Speed limit on the Road is 60 km/h and 110 km/h on the highway.

Time Taken = Distance/Time

Time taken on the road  $=\frac{\sqrt{30^2+x^2}}{60}$

Time taken on the highway $=\frac{50-x}{110}$

Total time taken to travel, T $=\frac{\sqrt{30^2+x^2}}{60}+\frac{50-x}{110}$

Minimum time taken occurs when the derivative of T equals 0.

$T^{'}=\frac{x}{60\sqrt{30^2+x^2}}-\frac{1}{110}\\\frac{x}{60\sqrt{30^2+x^2}}-\frac{1}{110}=0\\\frac{x}{60\sqrt{30^2+x^2}}=\frac{1}{110}\\110x=60\sqrt{30^2+x^2}$

Square both sides

$12100x^2=3600(30^2+x^2)\\12100x^2=3240000+3600x^2\\12100x^2-3600x^2=3240000\\8500x^2=3240000\\x^2=\frac{3240000}{8500} =381.18\\x=\sqrt{381.18} =19.52$

The road should be built such that it joins the highway at 19.52km from the point P.

In fact,

$|RM|=\sqrt{30^2+19.52^2}=35.79km$