An oil refinery is located 1 km north of the north bank of a straight river that is 3 km wide. a pipeline is to be constructed from the refinery to storage tanks located on the south bank of the river 5 km east of the refinery. the cost of laying pipe is $200,000/km over land to a point p on the north bank and $400,000/km under the river to the tanks. to minimize the cost of the pipeline, how far downriver from the refinery should the point p be located?

The point should be placed 3.362 km to the east of the refinery to minimize the cost.

Step-by-step explanation:

You should draw a graph that depicts the situation.

Let x be the distance from the refinery to the point P on the north bank in km and C the total cost of the pipeline.

According to the graph and using Pythagoras's theorem, the distance that the pipe should cover is given by

We know that the cost of laying pipe is $200,000/km over land to a point P on the north bank and $400,000/km under the river to the tanks. Therefore, the total cost of the pipeline is the cost of the part on land plus the cost of the amount underwater.

The domain of C(x) is [0, 5].

To find the minimum cost, we need to differentiate the cost function, set it to zero and solve for x to find the critical points.

Solving this equation with the help of graphing calculator (Desmos) we get that

We can use The Second Derivative Test to check if x is a minimum,

if C'(c) = 0 and C''(c) > 0, then C has a local minimum at c.

The second derivative is

Next, substitute into the second derivative

Because C''(3.362) > 0, then C(x) has a local minimum at 3.362.

Thus the point should be placed 3.362 km to the east of the refinery to minimize the cost.

college

47 - 90 = 43

43 - 180 = x

x = 137