Lagrange-Interpolation für die Kostenfunktion
Author: Prof. Dr. Dörte Haftendorn
Date: Nov 08

K_f sind die Fixkosten.

Kostenfunktion, ziehe an den Punkten

Ha 08

Kosten

BO...

BM

B = (1.0064633260711693, 2.3354393609295574)
K_f = Point(yAxis)
C = (2.5938343458096558, 2.675308641975308)
D = (3.5, 3)
K = Polynomial({K_f, B, C, D})
Kgrenz(x) = K'(x)
Kv(x) = K(x) - y(K_f)
stueckkv(x) = Kv(x) / x
A = Intersect(stueckkv, Kgrenz, (3.1594402358728417, 0.39293109268392623))
stueckk(x) = K(x) / x
E = Extremum(Kgrenz)
G = Intersect(stueckk, Kgrenz, (3.876417492508903, 0.8423603245377954))
BO: (x(G), 0)
F = Intersect(stueckkv, Kgrenz, (x(A), 1.14897))
BM: (x(F), 0)
lPug = Segment(G, BO)
kPug = Segment(F, BM)
a = true
b = true
Kgrenz'(x) = Kgrenz'(x)