Question

a power line is to be run to an offshore facility. the offshore facility is ( d = 3 ) 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 $90,000 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? (round your answers to two decimal places or the nearest cent if necessary, but do not round until your final computation.) length underground: miles length underwater: miles minimal cost: $

Explanation:

Step1: Define the cost function

Let the length of the power - line underground be \(y=(L - x)=(6 - x)\) miles (where \(x\) is the horizontal distance from the point on the shoreline closest to the offshore facility to the point where the underground line turns to go underwater). The length of the power - line underwater \(z=\sqrt{x^{2}+D^{2}}=\sqrt{x^{2}+9}\) miles.
The cost function \(C(x)=60000(6 - x)+90000\sqrt{x^{2}+9}\), \(0\leq x\leq6\).

Step2: Find the derivative of the cost function

Using the sum rule and the chain rule.
The derivative of \(y_1 = 60000(6 - x)\) is \(y_1^\prime=-60000\).
The derivative of \(y_2 = 90000\sqrt{x^{2}+9}=90000(x^{2}+9)^{\frac{1}{2}}\). By the chain rule \((u^n)^\prime=nu^{n - 1}u^\prime\), where \(u = x^{2}+9\), \(n=\frac{1}{2}\), \(u^\prime = 2x\). So \(y_2^\prime=90000\times\frac{1}{2}(x^{2}+9)^{-\frac{1}{2}}\times2x=\frac{90000x}{\sqrt{x^{2}+9}}\).
Then \(C^\prime(x)=-60000+\frac{90000x}{\sqrt{x^{2}+9}}\).

Step3: Set the derivative equal to zero and solve for \(x\)

\(-60000+\frac{90000x}{\sqrt{x^{2}+9}} = 0\).
First, move the constant term to the other side: \(\frac{90000x}{\sqrt{x^{2}+9}}=60000\).
Divide both sides by \(30000\): \(\frac{3x}{\sqrt{x^{2}+9}} = 2\).
Cross - multiply: \(3x = 2\sqrt{x^{2}+9}\).
Square both sides: \(9x^{2}=4(x^{2}+9)\).
Expand: \(9x^{2}=4x^{2}+36\).
Subtract \(4x^{2}\) from both sides: \(5x^{2}=36\).
Solve for \(x\): \(x^{2}=\frac{36}{5}\), \(x=\frac{6}{\sqrt{5}}\approx2.68\) (we take the positive value since \(x\geq0\)).

Step4: Calculate the length underground and underwater

The length underground \(y = 6 - x=6-\frac{6}{\sqrt{5}}\approx6 - 2.68 = 3.32\) miles.
The length underwater \(z=\sqrt{x^{2}+9}=\sqrt{\frac{36}{5}+9}=\sqrt{\frac{36 + 45}{5}}=\sqrt{\frac{81}{5}}=\frac{9}{\sqrt{5}}\approx4.02\) miles.

Step5: Calculate the minimal cost

\(C(x)=60000(6 - x)+90000\sqrt{x^{2}+9}\).
Substitute \(x = \frac{6}{\sqrt{5}}\) into the cost function:
\(C(\frac{6}{\sqrt{5}})=60000(6-\frac{6}{\sqrt{5}})+90000\sqrt{\frac{36}{5}+9}\)
\(=60000\times6-60000\times\frac{6}{\sqrt{5}}+90000\times\frac{9}{\sqrt{5}}\)
\(=360000+( - \frac{360000}{\sqrt{5}}+\frac{810000}{\sqrt{5}})\)
\(=360000+\frac{450000}{\sqrt{5}}\approx360000 + 201246.12=561246.12\)

Answer:

Length underground: \(3.32\) miles
Length underwater: \(4.02\) miles
Minimal cost: \(\$561246.12\)