More calculus help?

Question

Does anyone know how to do this problem:

An offshore oil rig well is 4 kilometers off the coast. The refinery is 9 kilometers down the coast. Laying pipe in the ocean is twice as expensive as on land. What path should the pipe follow in order to minimize the cost? Complete the following statement.

The pipe should be laid in the ocean from the rig to a point on the coast_______km from the refinery, then on land the remainder of the distance to the refinery.

Thanks for any help!

Answer 1

under the water, but not directly to the shore...

you need to optimize [ minimize here ] the cost, so find the equation of cost in terms of x...
let x = length of pipe under water, and then y = length on shore to the refinery...

then x^2 = 4^2 + a^2, since the pipe, shore and the point directly on the shore from the rig form a rt. triangle..... x = dist underwater, is the hypotenuse...

also a + y = 9,

then y = 9 - a, where a = sqrt ( x^2 - 4^2 ) is the distance along the coast, from a point directly across from the rig to the point the pipe comes to shore....

thus ... Cost = 2kx + 1ky, or C = 2kx + k ( 9 - sqrt ( x^2 - 16 ) )

k = cost of pipe in km, which is constant...

now take C ' = 0 and solve....

C ' = 2k - 2xk /[ 2 sqrt ( x^2 - 16 ) ] = 0
thus ... 2xk / [ 2 sqrt ( x^2 - 16 ) ] = 2k
x /[ 2 sqrt ( x^2 - 16 ) ] = 1 , or x = 2 sqrt ( x^2 - 16 ) ,... x^2 = 4 ( x^2 - 16 ), and solving... x = 8 / sqrt ( 3 ), or approx.... 4.62 km


then y = 9 - [sqrt( ( 64 / 3 ) - 16 ) ] = 9 - sqrt( 16 / 3) = 6.69 km from the refinery

by the first deriv test... we have a sign change for C ' ... x < 4.62 gives C ' < 0, and x > 4.62 gives C ' > 0, so x = 4.62 km gives us the minimum for Cost... , and y = 6.69 km is the distance along the shore that minimizes the Cost

also using the table function of a TI calculator on the Cost eqn... ignore k... I find the min cost at x = 4.618802... km ......
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be sure to choose someones work as the Best Answer ......

Answer 2

Let the required distance be x km. The length of pipe on land is x km and the length of pipe in the ocean is √((9-x)^2 + 4^2). Since ocean pipe costs twice as much to lay, the total cost will be a constant multiple of
f(x) = x + 2√((9-x)^2 + 16).
f'(x) = 1 + 2 (1/2) ((9-x)^2 + 16)^(-1/2) . [2(9-x) . (-1)]
= 1 - 2 (9-x) / √((9-x)^2 + 16)
= 0 <=> 2 (9-x) = √((9-x)^2 + 16)
=> 4 (9-x)^2 = (9-x)^2 + 16
<=> (9-x)^2 = 16/3
<=> 9-x = 4/√3 (since we know 9-x must be positive)
<=> x = 9 - 4/√3 ≈ 6.69.
We can check via first derivative test that this is a minimum of the cost function, as expected.