Optimization Problem

Question

Centerville is located at (9,0) in the xy-plane, Springfield is at (0,3), and Shelbyville is at (0,−3). The cable runs from Centerville to some point (x,0) on the x-axis where it splits into two branches going to Springfield and Shelbyville. Find the location (x,0) that will minimize the amount of cable between the 3 towns and compute the amount of cable needed. Justify your answer.

To solve this problem we need to minimize the following function of x:

f(x)=

We find that f(x) has a critical number at x=

To verify that f'(x) has a minimum at this criticial number we compute the second derivative f''(x) and find that its value at the critical number is: (positive number)

Thus the minimum length of cable needed is:

Answer 1

Since the point at which the cable splits is (x , 0), we need (9 - x) units of cable to lay along the x-axis. At that point, we need two hypotenuses of right triangles with legs = 3 and x, so by pthag thm c = √(x² + 9).

x: x-coordinate of split pt. f(x): total length of cable needed

f(x) = (9 - x) + 2√(x² + 9)

f^'(x) = - 1 + 2x / √(x² + 9) = 0

-√(x² + 9) + 2x = 0

4x² = x² + 9

x = √3

f^"(x) = (2√(x² + 9) - 2x²/√(x² + 9)) / (x² + 9)

f^"(√3) = 6√2 - √2 / 18 > 0

So we need f(√3) = 9 - √3 + 6√2 units of cable at a minimum.