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 50foot 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 stepbystep 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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