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 = (1.0864633260711694, 2.3554393609295574)
K_f = Point(yAxis)
C = (2.8900274348422506, 2.814742193173565)
la3(x) = (((x - x(K_f)) * (x - x(B))) * (x - x(C)))
D = (3.7873182633821716, 3.2417574437182277)
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, (3.2248578730009525, 0.41516818277161655))
stueckk(x) = K(x) / x
E = Extremum(Kgrenz)
G = Intersect(stueckk, Kgrenz, (4.01742575697962, 0.8512687625376134))
BO: (x(G), 0)
BM: (x(F), 0)
lPug = Segment(G, BO)
kPug = Segment(F, BM)
a = true
b = true
Kgrenz'(x) = Kgrenz'(x)