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 3 km east of the refinery. The cost of laying pipe is $300,000 per km over land to a point P on the north bank and $600,000 per km under the river to the tanks. To minimize the cost of the pipeline, how far from the refinery should P be located? (Give your answer correct to two decimal places.)

Let x be the distance between P and a point on the north bank directly opposite to the tanks.

The length of the pipe under water,
L1(x) = √(2²+x²).
The length of the pipe on land,
L2(x) = 3-x
Cost of laying the pipes,
C(x) = 600000 L1(x) + 300000 L2(x)

Differentiate C(x) with respect to x and equate the derivative to zero.
Solve for x.
The distance from the refinery to P is (3-x).
I get about 1.8 km.

To minimize the cost of the pipeline, we need to find the optimal location, P, on the north bank of the river. This is where the cost of laying pipe over land and under the river is minimized.

Let's denote the distance from the refinery to P as x km. So, the distance from P to the tanks will be (3 - x) km.

The cost of laying pipe over land from the refinery to P is given as $300,000 per km. Therefore, the cost of this segment will be 300,000x dollars.

The cost of laying pipe under the river from P to the tanks is given as $600,000 per km. So, the cost of this segment will be 600,000(3 - x) dollars.

The total cost of the pipeline will be the sum of the cost over land and the cost under the river:

Total cost = 300,000x + 600,000(3 - x)

To minimize the cost, we can take the derivative of the total cost with respect to x and set it equal to zero. Let's do that:

d(Total cost) / dx = 300,000 - 600,000

Setting this derivative equal to zero:

300,000 - 600,000 = 0

Simplifying:

-300,000 = 0

This equation has no solutions. It means that the cost is not minimized at any specific point; rather, the cost is minimized at the endpoints of the interval.

So, the optimal location, P, on the north bank is either at the refinery (x = 0 km) or at the south bank (x = 2 km). We need to calculate the cost for both cases and choose the minimum.

For x = 0, the cost is:

Total cost = 300,000(0) + 600,000(3 - 0) = 1,800,000 dollars

For x = 2, the cost is:

Total cost = 300,000(2) + 600,000(3 - 2) = 1,200,000 dollars

Comparing both costs, we find that the minimum cost is when P is located 2 km from the refinery, at the south bank of the river.

Therefore, the optimal location for P is 2 km from the refinery.

To minimize the cost of the pipeline, we need to find the point on the north bank where the cost of laying the pipe over the land plus the cost of laying it under the river is minimized.

Let's assume that the distance from the refinery to point P on the north bank is x km.

The cost of laying pipe over land will be $300,000 per km multiplied by x km, which gives us a cost of 300,000x dollars.

The distance from point P to the tanks on the south bank will be 3 km + x km (as the tanks are 3 km east of the refinery and we are considering x km from the refinery to point P).

The cost of laying the pipe under the river will be $600,000 per km multiplied by (3 + x) km, which gives us a cost of 600,000(3 + x) dollars.

The total cost will be the sum of the cost of laying pipe over land and under the river:

C(x) = 300,000x + 600,000(3 + x)

To minimize the cost, we need to find the value of x that minimizes the function C(x).

Next, we can differentiate the function C(x) with respect to x and set it equal to zero to find critical points:

C'(x) = 300,000 + 600,000 = 0

Simplifying the equation gives:

900,000 = 0

Since this equation has no solutions, there are no critical points.

Therefore, we can conclude that the minimum cost occurs at the endpoints of the interval, which are x = 0 (refinery) and x = 2 (opposite bank of the river).

To find the minimum cost, we can evaluate the function C(x) at these endpoints:

C(0) = 300,000(0) + 600,000(3 + 0) = 1,800,000 dollars

C(2) = 300,000(2) + 600,000(3 + 2) = 2,700,000 dollars

Comparing the costs, we see that the minimum cost occurs at point P located 2 km from the refinery on the north bank of the river.

Therefore, the point P should be located 2 km from the refinery.