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Math  /  Algebra

Question1 Points] DETAILS MY NOTES SCALCCC5 4.6.059.MI. PRACTICE ANOTHER
An oil refinery is located on the north bank of a straight river that is 2 km wide. A pipeline is to be constructed from the refinery to storage tanks located on the south bank of the river 10 km east of the refinery. The cost of laying pipe is $200,000/km\$ 200,000 / \mathrm{km} over land to a point PP on the north bank and $400,000/km\$ 400,000 / \mathrm{km} under the river to the tanks. To minimize the cost of the pipeline, how far (in km) from the refinery should PP be located? (Round your answer to two decimal places.) \qquad km Need Help? Read It Master It Submit Answer

Studdy Solution

STEP 1

1. The river is straight and 2 km wide.
2. The refinery is located on the north bank, and the storage tanks are 10 km east on the south bank.
3. The cost of laying pipe over land is 200,000/km200,000/km and under the river is 400,000/km400,000/km.
4. We need to find the point P P on the north bank that minimizes the total cost.

STEP 2

1. Define the variables and set up the cost function.
2. Express the cost function in terms of a single variable.
3. Find the derivative of the cost function.
4. Solve for the critical points to find the minimum cost.
5. Verify the solution and round to two decimal places.

STEP 3

Define x x as the distance from the refinery to point P P on the north bank.
The distance from P P to the point directly across the river from the storage tanks is 10−x 10 - x .
The total cost C(x) C(x) is the sum of the cost of the land pipe and the cost of the underwater pipe.

STEP 4

The cost of the land pipe is 200,000×x 200,000 \times x .
The underwater pipe forms the hypotenuse of a right triangle with legs 2 km (the width of the river) and 10−x 10 - x km (the horizontal distance), so its length is (10−x)2+22 \sqrt{(10-x)^2 + 2^2} .
The cost of the underwater pipe is 400,000×(10−x)2+22 400,000 \times \sqrt{(10-x)^2 + 2^2} .
The total cost function is:
C(x)=200,000x+400,000(10−x)2+4 C(x) = 200,000x + 400,000 \sqrt{(10-x)^2 + 4}

STEP 5

Find the derivative C′(x) C'(x) of the cost function C(x) C(x) .
C′(x)=200,000−400,000⋅10−x(10−x)2+4 C'(x) = 200,000 - 400,000 \cdot \frac{10-x}{\sqrt{(10-x)^2 + 4}}

STEP 6

Set C′(x)=0 C'(x) = 0 to find the critical points.
200,000=400,000⋅10−x(10−x)2+4 200,000 = 400,000 \cdot \frac{10-x}{\sqrt{(10-x)^2 + 4}}
Solve for x x :
12=10−x(10−x)2+4 \frac{1}{2} = \frac{10-x}{\sqrt{(10-x)^2 + 4}}
Square both sides and solve for x x :
14=(10−x)2(10−x)2+4 \frac{1}{4} = \frac{(10-x)^2}{(10-x)^2 + 4}
(10−x)2=14((10−x)2+4) (10-x)^2 = \frac{1}{4}((10-x)^2 + 4)
4(10−x)2=(10−x)2+4 4(10-x)^2 = (10-x)^2 + 4
3(10−x)2=4 3(10-x)^2 = 4
(10−x)2=43 (10-x)^2 = \frac{4}{3}
10−x=43 10-x = \sqrt{\frac{4}{3}}
x=10−43 x = 10 - \sqrt{\frac{4}{3}}
Calculate x x and round to two decimal places.

STEP 7

Calculate the numerical value:
x≈10−43≈10−1.1547≈8.8453 x \approx 10 - \sqrt{\frac{4}{3}} \approx 10 - 1.1547 \approx 8.8453
Round to two decimal places:
x≈8.85 x \approx 8.85
The point P P should be located approximately 8.85 \boxed{8.85} km from the refinery.

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