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
we know at (40,1) slope = .06
so.. .06 = 4800a + 80b + c
4800a + 80b = .02 (equ#2)
I then put c=.04 into equ#1, and solved this with equ#2 to get
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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