calculus question

Question

Two isolated farms are situated 12km apart on a straight country road that runs parallel to the main highway 20km away. The power company decides to run a wire from the highway to the junction box, and from there, wires of equal length to 2 houses. Where should the junction box be placed to minimize the length of wire needed?

Help!?

Answer 1

Set up a coordinate system where one farm is at (-6,0) and the other at (6,0), which is 12 km apart, and the highway is along the line y = 20.  The box will be along the perpendicular bisector of the line joining the 2 farms, at (0,y), with 0 ≤ y ≤ 20, and y to be determined.  I assume you want to minimize the total length of the wire, which includes the perpendicular length of 20 - y from the highway to the junction box, plus the length of the hypotenuses from the junction box to the farms at (-6,0) and (6,0).  The hypotenuses will have the same length.  Then the total length as a function of y is

 

L(y) = (20 - y) + 2√(y² + 6²) = (20 - y) + 2√(y² + 36)

 

Set dL/dy = 0 and solve for y.

 

dL/dy = -1 + 2y / √(y² + 36) = 0

 

2y / √(y² + 36) = 1

 

4y² = y² + 36

 

3y² = 36

 

y² = 36/3 = 12

 

y = √12 = 2√3 ≅ 3.464 km from the road connecting the farms, toward the highway. 

 

Note that if you place the junction box at the highway, it will be at (0,20).  Then there is no wire length from the highway to the junction box, and the total length is just the sum of the 2 hypotenuses.

 

L = 2√(20² + 6²) ≅ 41.76

 

But if you place the junction box between the highway and farm road at (0,√12),

 

L = (20 - √12) + 2√[(√12)² + 6²] ≅ 30.39,

 

which is shorter.

Answer 2

IMHO this does not appear to be a calculus problem.

Draw a diagram.

The box should be at the intersection of the highway and the perpendicular bisector of the farm road.