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November 28, 2014

November 28, 2014

Posted by **John** on Wednesday, March 17, 2010 at 1:48pm.

- Calculus Optimization -
**Reiny**, Wednesday, March 17, 2010 at 2:25pmDid you make a diagram?

Mine has I as the island, O as the point on shore opposite the island, and T as the transformer station.

So IO = 6 km and OT = 12 km

Let P be any point between O and T,

let OP = x, then PT = 12-x

Cost = 6000(IP) + 4000(12-x)

IP^2 = x^2 + 6^2

IP = (x^2 + 36)^(1/2)

Cost = 6000(x^2 + 36)^(1/2) + 48000 - 4000x

d(Cost)/dx = 3000(x^2+36)^(-1/2)(2x) - 4000

= 0 for a max/min of Cost

2x(3000)/√(x^2 + 36) = 4000

3x/√(x^2 + 36) = 2

square and cross multiply

9x^2 = 4x^ + 144

5x^2 = 144

x= √144/5 = 5.3667

Form your word conclusion

- Calculus Optimization -
**Damon**, Wednesday, March 17, 2010 at 2:31pmsay it goes 12-x km down the shoreline, then heads offshore straight for the island. The resulting path length is 12-x + the hypotenuse sqrt(36 + x^2)

cost = (12 -x)(4000) + [(36+x^2)^.5]6000

0 =dc/dx = -4000 + .5 [ (36+x^2)^-.5 ] (2x)(6000)

which is

4000 = 6000x/sqrt(36+x^2)

sqrt(36+x^2) = (3/2)x

36+x^2 = (9/4)x^2

36 = (5/4)x^2

x^2 = 28.8

x = 5.37 km

12-x = 6.63 miles on shore then head under water

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