How far down the river should the point P be located?

Question

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 8 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 downriver from the refinery should the point P be located? (Round the answer to two decimal places.)

Answer 1

Start out by drawing out the scenario. Here is my diagram:

https://mail.google.com/mail/?ui=2&ik=aa7b77c4f8&view=att&th=1305e5605f822138&attid=0.1&disp=inline&zw



Suppose that we construct a pipeline across the shore that x km away from the refinery. By the Pythagorean Theorem, we see that the length of pipe across the river is:

L(pipe,river) = √(x^2 + 3^2) = √(x^2 + 9).



Then, the length of the remaining pipe across the bank is:

L(pipe,bank) = 8 - x.



The total cost of the pipeline is:

C = 600000*L(pipe, river) + 300000*L(pipe, bank)

= 600000√(x^2 + 9) + 300000(8 - x).



By differentiating:

dC/dx = 600000x/√(x^2 + 9) - 3000000

= 300000[2x - √(x^2 + 9)]/√(x^2 + 9), by combining fractions.



By setting dC/dx = 0:

2x - √(x^2 + 9) = 0 ==> x = √3.



You can show that this produces a minimum by showing that d^2C/dx^2 > 0.



Therefore, the point P should be located √3 ≈ 1.73 km away from the refinery.



I hope this helps!

Answer 2

Let P be located x km east of the refinery. By the Pythagorean Theorem, the distance d from the refinery to P is:



d(x) = √(x^2 + 1^2)

= √(x^2 + 1)



And, also by the Pythagorean Theorem, the distance k from P to the storage tanks is:



k(x) = √((8 - x)^2 + 3^2)

= √(x^2 - 16x + 73)



The total cost C of the pipeline is given as:



C(x) = 300,000d + 600,000k

= 300,000√(x^2 + 1) + 600,000√(x^2 - 16x + 73)



And, the minimum cost of the pipeline will be given for dC/dx = C'(x) = 0.



C'(x) = 0

=>

300,000x / √(x^2 + 1) + 300,000(2x - 16) / √(x^2 - 16x + 73) = 0

=>

x / √(x^2 + 1) = (16 - 2x) / √(x^2 - 16x + 73)

=>

x√(x^2 - 16x + 73) = (16 - 2x)√(x^2 + 1)

=>

x^2(x^2 - 16x + 73) = (16 - 2x)^2 (x^2 + 1)

=>

x^4 - 16x^3 + 73x^2 = 4x^4 - 64x^3 + 260x^2 - 64x + 256

=>

3x^4 - 48x^3 + 187x^2 - 64x + 256 = 0



This looks really hard to solve. But, using a calculator, we discover that:



x ≈ 6.296361114

= 6.30 km (rounded to two decimal places)

Answer 3

4.85 km is unlikely to be correct as you could reduce costs just by placing P 3km east of the oil refinery (though this is unlikely to be the right position of P). This question is actually quite long. Let x be the distance of P from the refinery, let the cost be c and to simplify the figures let the cost be $1 per km on land and $2 under water (this won't make any difference to the problem). The question is to find the x that minimizes c where c = x+2sqrt[ (3-x)^2+2^2 ] (use pythagoras to work out the distance under water) take x to the other side and square to get (c-x)^2 = 4[ (3-x)^2+2^2] expand and simplify -3 x^2+(24-2c)x+(c^2-52) = 0 This quadratic equation only has a solution when it's discriminant is >=0 The discriminant (B^2-4AC, for the equation AX^2+BX+C=0) is (24-2c)^2+12(c^2-52) which simplifies to 16c^2-96c-48 The smallest positive c which makes the above positive is when c=3+2 sqrt(3) (use the quadratic formula), at this value of c the discriminant is 0. Substitute this value of c into -3 x^2+(24-2c)x+(c^2-52) = 0 to get x = 3- 2 sqrt(3)/3 = 1.84km

Answer 4

extremely tough factor. browse onto the search engines. it could help!