Finding cheapest route vs distance.?

Question

Z
|
|
|
A-----------------O----

Z to A =10km
A to O=16km

Combination of horizontal line and triangle. Aim is to get a line from O to Z, but here's the thing. It has to cost as least as possible. Each km off the horizontal axis is 4/3 expensive as any km on the horizontal axis. What's the cheapest route? Please show all calculations/methods you use.

Answer 1

Let the distance moved horizontally along OA to B say be 16 - x so that AB = x.



BZ^2 = 10^2 + x2 ----> BZ = sqrt(100 + x^2)



Cost of moving along BZ = (4/3)*sqrt(100 + x^2)



Total cost T = 16 - x + (4/3)*sqrt(100 + x^2)



Differentiate tio find dT/dx and equate to zero in the usual way. Solve the equation to find the value of x which minimises total cost.



Can you finish with that help? You should find that the answer is to move between 4.5 and 5 km along OA before heading straight for X.

Answer 2

Let x be a point on the horixontal axis between A and O

Route is to be a straight line From O to X, then X to Z



Cost (c) = distance from Z to x * 4/3 + distance from x to O



C(x) = 4/3 * sqrt(10^2 + x^2) + (16 - x)



. . . . C is mimimized when it's derivative = 0



C ' (x) = (4x) / (3 sqrt(x^2 + 100)) - 1



(4x) / (3 sqrt(x^2 + 100)) - 1 = 0



x = 30 / sqrt(7)



x is about 11.34 km east of A

and 4.66 km west of O

Answer 3

OZ^2 = 16^2 + 10^2



OZ^2 = 356



OZ = \/356 = 18.868 km ANSWER