Doccier fundamentos de matemáticas

Page 1

DOSSIER INFORMATIVO INTEGRADOR FUNDAMENTOS DE MATEMATICAS

KELLY LAZARO LOPEZ CODIGO 221309 KAREN ALVERNIA CODIGO 221339 CHRISTIAN CORTES S CODIGO 230801

DOCENTE: MARIA DEL PILAR PEREZ

UNIVERSIDAD FRANCISCO DE PAULA SANTANDER OCAÑA FACULTAD DE CIENCIAS ADMINISTRATIVAS Y ECONOMICAS CONTADURIA PÚBLICA NOCTURNO TERCER SEMESTRE 2015


X₂=30 PARA GRAFICAR: R(X)= -20000X² + 1200000X R (30)= -20000 (30)² + 1200000 (30) R (30)= 18000000 FUNCION DE INGRESO R(X)= -20000X² + 1200000X A= -20000

b=1200000

c=0

Xv =

Xv =

=

Xv= 30

RV = R (30) = -20000(30)² + 1200000 (30) RV = 18000000 Pv = (30, 18000000) PUNTOS DE CORTE:


X₁ = 0 √

X₂ = 60 30000000

25000000

20000000

30, 18000000 R(x)

15000000

C(x) 10000000

5000000

0 0

COSTOS

10

20

30

C(x) = 600000X

Como el X= 30 es el punto de equilibrio C (30) = 18000000

40

50

60

70


Xmax=

Xmax=

Xmax=15 µ(x) = -20000x² + 600000x µ (15) =-20000(15)² + 600000(15) µ(x) =8700000

Punto de equilibrio µ(x)=0

R(x)=C(x)

-20000x²+1200000x=600000x -20000x²+1200000x-600000x=0 -20000x²+600000x=0 a=-20000

Xᵢ= X₁= X₁= X₁=0

b=600000

c=0


(10, 1000000) (20800000) (µ)=600000

P-p˳=

(X - X˳)

P-1000000 = P-100000 =

(X - 10) (X - 10)

P-1000000= -20000 (X – 10)

P=-20000X + 200000 + 1000000 P=-20000X + 1200000

‫ ٭‬Ƈ (X)= Ƈ(µ) . X

Ƈ (X) = 600000X ‫ ٭‬R(X)=

P.X

R(X) = [-20000X + 1200000] X R(X) = -20000X² + 1200000X ‫ ٭‬µ (X) = R (X) – Ƈ (X)

µ (X) = -20000X² + 1200000X - 600000X µ (X) = -20000X² + 600000X función utilidad a= -20000

b= 600000


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