Optimization of Road Repair and Construction Project

[storm13]

Homework Statement

Albany is 12km north of Rochedale. Bells Creek is 5km west of Albany. The road is to be repaired and repairs cost $96 000 per km. The cost of laying a new road is $120 000 per km.

It has been decided to repair the old road from Rochedale as far as point K, and then to build a new road from K to Bells Creek. Find the length of the new road such that the cost of the whole project is a minimum.

Homework Equations

The Attempt at a Solution

Let C = total cost
Let the distance from point K to Bells Creek = x
Let the distance from Rochedale to point K = y
Let the distance from point K to Albany = b

Thus the y can be written in terms of b using the 12km from the initial question.
y = 12 - b

C = 120000x + 96000y

Substitute y=12 - b into the equation

C = 120000x - 96000b + 1152000

Then there is a right angle triangle from point K to Albany to Bells Creek, with the hypotenuse being the new road and the variable x defined above.

Thus the side b (dsitance from point K to Albany) can be written in terms of x

b = √(x^2 - 25)

Then substitute the value for b into the total cost equation.

C = 120 000x - 96000(√(x^2 - 25)) + 1152000

Not really sure where to go from here. I have done lots of problems and understand the concept of finding the minimum or maximum by making the derivative equal to zero but am not sure how to go about either simplifying or deriving the middle term because of the square root sign.

Any help would be greatly appreciated.

 

[jedishrfu]

are you stuck on taking the derivative of the eqn?

use the chain rule with u = (x^2 -5) so that du/dx = 2*u and for the term then --> 1/2* u^-3/2 * 2*u

 

[storm13]

Well if i use the chain rule to differentiate i think this is what i get:

(for the middle part):

=1/2 . 96000(x^2 - 25)^-1/2 . (2x) [Note using (.) to mean multiplication]
=48000(x^2 - 25)^-1/2 . (2x)
= 96000x(x^2 - 25)^-1/2

So the derivative of the total coat is:

C' = 120 000 - 96000x(x^2 - 25)^-1/2

Then i make this equal to zero to solve for the minimum

0 = 120 000 - 96000x(x^2 - 25)^-1/2

120000 = 96000x(x^2 - 25)^-1/2
120/96 = x(x^2 - 25)^-1/2

This is where i get stuck i don't know how to solve the rest of it (this is assuming the part before is correct??)

 

[eumyang]

0 = 120000 - 96000x(x^2 - 25)^-1/2

Assuming that this is correct, multiply both sides by √(x² - 25):
0 = 120000\sqrt{x^2 - 25} - 96000x
Isolate the square root, and then square both sides. I'll let you figure out the rest.

 

[storm13]

Thank you for that help.

This is what i got for the rest.

0 = 120 000√(x^2 - 25) - 96 000x
96000x = 120 000√(x^2 - 25)
x = 1.25√(x^2 - 25)
x^2 = 1.5625(x^2 - 25)
x^2 = 1.5625x^2 - 39.0625
x^2 + 39.0625 = 1.5625x^2
39.0625 = 0.5625x^2
69.44 ≈ x^2
8.333 ≈ x

Is my working correct? I have checked the sign of the derivative before and after this x value and it is negative before and positive after so it appears to be a minimum.

Then to finish the problem,

b = √(8.3333^2) - 25
b =√44.44
b ≈ 6.66

thus y = 12 - 6.66
y = 5.34

C = (120000 x 8.33) + (96000 x 5.34)
C = 999600 + 512640
C = 1 512 240

Thus total cost at a minimum is $1 512 240.

Thank you both for replying. I appreciate it.

 

[eumyang]

Thank you for that help.

This is what i got for the rest.

0 = 120 000√(x^2 - 25) - 96 000x
96000x = 120 000√(x^2 - 25)
x = 1.25√(x^2 - 25)
x^2 = 1.5625(x^2 - 25)
x^2 = 1.5625x^2 - 39.0625
x^2 + 39.0625 = 1.5625x^2
39.0625 = 0.5625x^2
69.44 ≈ x^2
8.333 ≈ x

Is my working correct?

Well, when take the square root of both sides, you should at first include both the positive and negative roots. So x ≈ ±8.333. But of course, x ≈ -8.333 is an extraneous solution for the equation 0 = 120000√(x^2 - 25) - 96000x, and x represents the distance from point K to Bells Creek, which can't be negative anyway.

 

[storm13]

Thanks again. Appreciate it.