Coastal Cable Route with Endpoint and Cost Sensitivity Checks
A coastal council must connect an offshore monitoring platform A to a treatment plant C on a straight shoreline. The point B on the shore nearest A is 8 km from A, and C is 15 km along the shore from B. An underwater cable costs $1.2 million per kilometre; a land cable costs $0.5 million per kilometre. The cable will run straight from A to a junction D on the shoreline, then straight along the shore from D to C. Let x be the distance from B to D in kilometres, with 0≤x≤15. The council's engineer proposes choosing x by setting the derivative of cost to zero and rounding to two decimal places. A councillor objects that the cheapest route might simply use the shortest underwater run or no underwater run at all. You must decide the optimal x, compare endpoints, and advise how the decision changes if land cable prices rise. All distances are in kilometres and costs in millions of dollars.
1. Write the total cost C(x) and find the critical point. Give both an exact form and a decimal value.
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The underwater length is sqrt(x^2+8^2)=sqrt(x^2+64), and the land length is 15-x. Therefore C(x)=1.2sqrt(x^2+64)+0.5(15-x), 0≤x≤15. Differentiate: C'(x)=1.2x/sqrt(x^2+64)-0.5. Set C'(x)=0: 1.2x/sqrt(x^2+64)=0.5. Squaring is safe here because x≥0 and both sides are non-negative: 1.44x^2=0.25(x^2+64), so 1.19x^2=16, x^2=16/1.19=1600/119, and x=40/sqrt119 ≈ 3.67 km.
2. Justify that this critical point gives the global minimum on 0≤x≤15, not merely a stationary point.
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C'(x)=1.2x/sqrt(x^2+64)-0.5. The function g(x)=1.2x/sqrt(x^2+64) is strictly increasing for x≥0, because its derivative is positive. Hence C'(x) is negative before the critical point and positive after it, so C decreases then increases; the critical point is the unique global minimum on the interval. Endpoint values confirm this: C(0)=1.2(8)+0.5(15)=9.6+7.5=17.1, C(15)=1.2sqrt(225+64)+0=1.2(17)=20.4. At x=40/sqrt119, sqrt(x^2+64)=sqrt(1600/119+64)=sqrt(9216/119)=96/sqrt119. Thus C=1.2(96/sqrt119)+0.5(15-40/sqrt119) =115.2/sqrt119+7.5-20/sqrt119 =7.5+95.2/sqrt119 ≈ 7.5+8.727=16.227. This is lower than both endpoints, so the interior junction is optimal.
3. With the underwater rate still $1.2 million/km, find the largest land-cable rate for which an interior junction below C can be optimal, and explain the endpoint case when the land rate exceeds that threshold.
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For land rate c, C'(x)=1.2x/sqrt(x^2+64)-c. An interior critical point requires c=1.2x/sqrt(x^2+64). On 0≤x≤15 the right side increases from 0 to its maximum at x=15: 1.2(15)/sqrt(225+64)=18/17 ≈ 1.059. So an interior critical point with 0<x<15 exists for 0<c<18/17. At c=18/17 the stationary point has moved to the endpoint x=15. For c>18/17, C'(x)<0 throughout 0≤x≤15, so cost decreases all the way to x=15. That route puts the junction at C: the cable runs directly underwater from A to C, with no land section. Thus the threshold is 18/17 million dollars per kilometre; beyond it, the endpoint x=15 is optimal.