Extremwertaufgabe: maximales Schachtelvolumen
Author: T. Warncke (nach Jens Helling)

Schiebe den farbigen Punkt des grünen Schiebereglers um die Höhe der
Schachtel zu variieren.

Höhe der Schachtel =

cm

cm^3

Von der Schachtel-Höhe h abhängiges

Volumen der Schachtel

V(h)=

a = 3 (slider from 0 to 4.999, step 0.01)
B: (-10 + 2 a, 0)
C: (x(B), a)
D: (0, a)
c = Segment(D, C)
F = Point(Circle[D, 10 - 2 a])
b = Segment(D, F)
α = Angle(F, D, C)
E: (x(F), y(F) - a)
A = (0, 0)
d = Segment(E, A)
G: (x(E) - 10 + 2 a, y(E))
H: (x(G), y(G) + a)
e = Segment(E, G)
f = Segment(F, H)
g = Segment(H, C)
h = Segment(G, B)
i = Segment(C, B)
Breite = Segment(B, A)
Höhe = Segment(A, D)
l = Segment(F, E)
m = Segment(H, G)
I = Intersect(e, g)
J = Intersect(c, d)
P, j_1, e_1, i_1, c_1 = Polygon(J, E, I, C)
n = Segment(I, E)
p = Segment(E, J)
q = Segment(E, A)
r = Segment(G, E)
f_1(x) = (4 x^(3) - 40 x^(2) + 100 x) 0.1
g_1 = Function(f_1, 0, 5)
v: (a, g_1(a))
V: y(v 10)
T1: V + ""
T3: a + ""