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 = (3.0718228031953534, 2.919317356572258)
la3(x) = (x - x(K_f)) * (x - x(B)) * (x - x(C))
D = (3.4824136129074414, 3.2591866376180096)
c3: y(D) / la3(x(D))
la0(x) = (x - x(B)) * (x - x(C)) * (x - x(D))
c0: y(K_f) / la0(x(K_f))
la1(x) = (x - x(K_f)) * (x - x(C)) * (x - x(D))
c1: y(B) / la1(x(B))
la2(x) = (x - x(K_f)) * (x - x(B)) * (x - x(D))
c2: y(C) / la2(x(C))
K(x) = c0 * la0(x) + c1 * la1(x) + c2 * la2(x) + c3 * la3(x)
Kv(x) = K(x) - y(K_f)
stueckkv(x) = Kv(x) / x
Kgrenz(x) = K'(x)
F = Intersect(stueckkv, Kgrenz, (2.7379031970361183, 0.41944253570983364))
stueckk(x) = K(x) / x
E = Extremum(Kgrenz)
G = Intersect(stueckk, Kgrenz, (3.3936094694519006, 0.9347525138181759))
BO: (x(G), 0)
BM: (x(F), 0)
lPug = Segment(G, BO)
kPug = Segment(F, BM)
a = true
b = true
Kgrenz'(x) = Kgrenz'(x)