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# In dire need of help to solve calculus problem!?

Centerville is the headquarters of Greedy Cablevision Inc. The cable company is about to expand service to two nearby towns, Springfield and Shelbyville. There needs to be cable connecting Centerville to both towns. The idea is to save on the cost of cable by arranging the cable in a Y-shaped configuation. Centerville is located at (9,0) in the xy-plane, Springfield is at (0,7), and Shelbyville is at (0,- 7). 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.

Thank you!

### 1 Answer

- 1 decade agoFavorite Answer
Heh, had to draw a picture for this ^^; At least it's more fun than the math problem I'm struggling with >.<

Anyway the total amount of cable, say, d is =sqt(x^2+49)+sqt(x^2+81)+9-x.

So we want to minimize the cable, meaning we take the derivitive.

That's x/sqt(x^2+49)+x/sqt(x^2+81)-1=0

My I like those numbers.

x/sqt((x+7)^2)+x/(sqt(x+9)^2)=1

x/(x+7)+x/(x+9)=1

x(x+9)+x(x+7)=x^2+16x+63

x^2+9x+x^2+7x=x^2+16x+63

2x^2+16x=x^2+16x+63

x^2=63

x=rad(63)

x=3rad(7)

Pretty sure that's the answer, that's certainly the methodology, but I'd check my numbers if I were you.