Posted by **Jon** on Monday, December 3, 2007 at 10:41pm.

A section of highway connecting two hillsides with grades of 6% and 4% is to be build between two points that are separated by a horizontal distance of 2000 feet. At the point where the two hillsides come together, there is a 50-foot difference in elevation.

a) Design a section of highway connecting the hillsides modeled by the function f(x) = ax^3 + bx^2 + cx + d (-1000 less than or equal to x less than or equal to 1000). At the points A and B, the slope of the model must match the grade of the hillside.

b) Use a graphing utility to graph the model.

c) Use a graphing utility to graph the derivative of the model.

d) Determine the grade at the steepest part of the transitional section of the highway.

I need to show work step-by-step for this, so please format your answer like that. Thanks! :)

- Calculus -
**Reiny**, Tuesday, December 4, 2007 at 12:00am
I started by putting point A at the origin (0,0), then point B would be at (2000,50)

I realized I was going to work with very large numbers, so I "scaled" my graph back by a factor of 50:1

so point A was still (0,0) but point B became (40,1)

let f(x) = ax^3 + bx^2 + cx + d

at (0,0) this would give me d = 0

so f(x) = ax^3 + bx^2 + cx

I then subst. (40,1) into that to get

1 = 64000a + 1600b + 40c (equ#1)

also f'(x) = 3ax^2 + 2bx + c

we know at (0,0) slope = .04

so ....c=.04

we know at (40,1) slope = .06

so.. .06 = 4800a + 80b + c

but c=.04

4800a + 80b = .02 (equ#2)

I then put c=.04 into equ#1, and solved this with equ#2 to get

a=1/32000

b=-13/8000

c=1/25

then finally

f(x) = (1/32000)x^3 - (13/8000)x^2 + (1/25)x

I set the derivative of that equal to zero, there was no real solution, so there is no max/min to this function.

there is a point of inflection at x=17.3

which translates into 17.3*50 = 865 m horizontal from A

My guess is that the slope of .06 is the largest in your domain

Please check my work, hard to do arithmetic like this while watching a football game, lol

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