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Mathematik Klausurübungen Lösung

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Sommersemester Mathematik Lösungen Übungen für Klausur Mathematik Klausurübungen Lösung

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  • July 22, 2023
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LÖSUNGEN ZU DER PROBEKLAUSUR MATHEMATIK II



Aufgabe 1.
Matrix A + B AT · B T C T · B (AT 4B) · C A · B · C
a)
Typ nd (1,1) nd (1,3) (4,3)

Aufgabe 2. Die Lösung ist M1 = 10, M2 = 20 und M3 = 10.

Aufgabe 3.
0 1
✓ ◆ 1 2 3 ✓ ◆
2 3 4 1 0 1
a) MRZ = , MZE = @1 2 1 A , MRE =
5 2 1 0 2 0
1 0 1
✓ ◆
10 10 14
M=
8 16 18
✓ ◆
720
b) r =
940
c) Rohmaterialkosten der Zwischenprodukte = 10 20 · MRZ = 120 70 60 , also ist es
günstiger, die Zwischeprodukte einzukaufen.

Aufgabe 4.
a) 0  x  26.
b) E(x) = 260x 10x2 , k(x) = 440 x + 20, k 0 (x) = 440
x2
0
c) k < 0 für alle x 2 [0, 26], also sind die Durchschnittskosten streng monoton fallend im
ganzen Definitionsbereich mit Minimum in x = 26.
d) G(x) = 10x2 + 240x 440.
Break-Even-Points: 10x2 + 240x 440 = 0 ) x1 = 2, x2 = 22. Die Parabel ist nach
unten geö↵net, also ist der Gewinn negativ für die Outputmengen x 2 [0, 2] [ [22, 26].
e) E 0 (x) = 20x + 260 = 0 , x = 13, E 00 (x) = 20 < 0 für alle x, also liegt in x = 13 ein
globales Maximum vor mit dem Wert E(13) = 1690 GE.

Aufgabe 5.
2
a) p(m) = 3m + 32.
2
b) ✏p (m) = 2
3
m+32
m.
3
✏p (21) = 79 . Die Preis reagiert unelastisch auf eine Änderung bei einer Nachfragemenge
21. Eine 1%-ige Erhöhung der Nachfragemenge führt zu einer 0, 77%-igen Senkung des
Preises.
p
c) m = 2%1% = 2 = ✏p (m)
2m
2= 2
3
m+32
() m = 32.
3
Bei einer Menge von 32 ist das der Fall.

Aufgabe 6.
p 1 4
3
a) xr1 (r1 , r2 ) = r13 4 r2 + ln r2 , xr2 (r1 , r2 ) = 16 r1 r2 4 + r1 r12 .
b) dx(r1 , r2 ) = xr1 (r1 , r2 )dx + xr2 (r1 , r2 )dr2 with dr1 = 1 and dr2 = 2.
dx(3, 16) = 55, 13. Die Produktion fällt um etwa 55,13 ME.
1

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