Calculus problem!!?

Question

An oil refinery is located on the north bank of a straight river that is 1 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 $500,000 per km over land to a point P on the north bank and $1,000,000 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.)

Answer 1

See my sketch: http://i24.tinypic.com/258c95k.gif

(It would be very hard for me to try to solve a problem like this without a sketch. I hope you made a stab at a sketch as well.)

Based on the lengths as indicated in the sketch,

Cost of land pipeline = (500,000)(5-x)
Cost of underwater pipeline = (1,000,000)[sqrt(1 + x²)]
Total cost is the sum of these two. So

C(x) = (500,000)(5-x) + (1,000,000)[(1 + x²)^½], 0 ≤ x ≤ 5

Minimize C.

Technically, you should consider C(0) and C(5) as well as any x in (0,5) for which C' = 0.

Note that x is not the distance from the refinery to P. I chose to use x for the length that I did because it makes the algebra easier.

Answer 2

Let
length of pipeline over land = x km
length of pipeline under water = z km

But by pythagoras theorem
z^2 = (5 - x)^2 + 1^2
z^2 = 26 - 10x + x^2
z = (26 - 10x + x^2)^1/2

Total cost is given by
C = 5 * 10^5 x + 1 * 10^6 z
C = 5 * 10^5 x + 1 * 10^6 * (26 - 10x + x^2)^1/2
differentiate both side wrt to x
dC/dx = 5*10^5 + 1 * 10^6 *(-10 + 2x) / (26 - 10x + x^2)^1/2
To find minimum dC/dx = 0
0 = 5*10^5 + 1 * 10^6 *(-10 + 2x) / (26 - 10x + x^2)^1/2
-5*10^5(26 - 10x + x^2)^1/2 = 1 * 10^6 *(-10 + 2x)
-(26 - 10x + x^2)^1/2 = 2(2x - 10)
square both side, we get
(26 - 10x + x^2) = 4(2x - 10)^2
(26 - 10x + x^2) = 4(4x^2 - 40x + 100)
26 - 10x + x^2 = 16x^2 - 160x + 400
15x^2 - 150x + 374 = 0
x = (150 +- sqrt(150^2 - 4* 15 * 374))/2*15
x = (150 +- sqrt(60))/30
x = (150 +- 7.75)/30
either x = 5.26 or x = 4.74
x cannot be 5.26 because x has to be less than 5, so x is 4.74

P should be at 4.74 km away from refinery

Total cost = 5 * 10^5 x + 1 * 10^6 * (26 - 10x + x^2)^1/2
= 5*10^5(4.74 + 2 *(26 - 10*4.74 + 4.74^2)^1/2)
= 5*10^5(4.74 + 2.06)
= 34 * 10^5