Questions: A power line is to be run to an offshore facility as shown in the figure above. The offshore facility is D=1 miles at sea and L=6 miles along the shoreline from the power plant. It costs 60,000 per mile to lay a power line underground and 95,000 per mile to run the line underwater. How much of the power line should be run underground, and how much underwater to minimize the overall costs? Length underground (in miles): Length underwater (in miles): Minimal cost (in dollars):

Solution Steps

Let $x$ be the distance along the shoreline where the power line enters the water. The underground length is then $6 - x$ miles. The underwater length is given by the Pythagorean theorem: $\sqrt{1^2 + x^2} = \sqrt{1 + x^2}$ miles. The cost function is $C(x) = 60000(6 - x) + 95000\sqrt{1 + x^2}$.

To minimize the cost, we need to find the critical points of $C(x)$ by taking its derivative and setting it to zero. $C'(x) = -60000 + 95000 \frac{x}{\sqrt{1 + x^2}} = 0$ $95000x = 60000\sqrt{1 + x^2}$ Squaring both sides: $9025 \times 10^6 x^2 = 3600 \times 10^6 (1 + x^2)$ $9025x^2 = 3600 + 3600x^2$ $5425x^2 = 3600$ $x^2 = \frac{3600}{5425} = \frac{144}{217}$ $x = \sqrt{\frac{144}{217}} \approx 0.817$ miles. Since the shoreline has length 6 and we chose x to be the distance to where the line goes underwater, this value makes sense.

Length underground: $6 - x = 6 - 0.817 = 5.183$ miles. Length underwater: $\sqrt{1 + x^2} = \sqrt{1 + \frac{144}{217}} = \sqrt{\frac{361}{217}} \approx 1.286$ miles. Minimal cost: $60000(5.183) + 95000(1.286) = 310980 + 122170 = \$433150$

Final Answer: