micro-economia: geometría una caisha de costat quadrat

F(x,z) = 16·( p( 2x^{2}+4xz )+(-1)mx^{2}z )


d_{x}[F(x,z)] = p( 4x+4z )+(-1)(2m)xz
d_{z}[F(x,z)] = p( 4z )+(-1)mx^{2}


x = 0
z = 0


p=(m^{2}/16)


d_{x}[F(x,z)] = (m^{2}/4)( x+z )+(-1)(2m)xz
d_{z}[F(x,z)] = (m^{2}/4)( z )+(-1)mx^{2}


x=(m/4)
z=(m/4)

índex de etiquetes

index de etiquetes.

economia: modelo de línea de la ecuación de primer orden de logaritmo no inverso

f(x) = e^{(m/x)}
ln( f(x) ) = (m/x)


d_{z}[h(z)] + (-1)·( x^{2}/m )·h(z) = 0


h(z) = e^{(x^{2}/m)·z}


( ln(e^{(x^{2}/m)·(m/x)}) ) = x


( ln(e^{(x^{2}/m)·ln( f(x) )}) ) = x


( ln( h( ln(f(x)) ) ) ) = x


m = 3 & x = (3/4)·m ==> ln(f(3/4)) = 4€


d_{z}[h(z)] + (-1)·(3/16)·h(z) = 0


ln(h(4) ) = (3/4)€


ln(f(3/4))·ln(h(4) ) = 3€


m = 5 & x = (1/10)·m ==> ln(f(1/10)) = 50€


d_{z}[h(z)] + (-1)·(1/500)·h(z) = 0


ln(h(50) ) = (1/10)€


ln(f(1/10))·ln(h(50) ) = 5€

economia: modelo de superficie de la ecuación diferencial de primer orden

f(x) = e^{(m/x^{2})}
ln( f(x) ) = (m/x^{2})


d_{y}[g(y)] + (-1)·( m/x^{3} )·g(y) = 0


g(y) = e^{(m/x^{3})·y}


(1/ln(e^{(m/x^{3})·(x^{2}/m)})) = x


(1/ln(e^{(m/x^{3})·( 1/ln( f(x) ) )})) = x


( 1/ln( g( 1/ln(f(x)) ) ) ) = x


m = 2 cajas & x = 20 cm ==> ( 1/ln( f(20) ) ) = 200€


d_{y}[g(y)] + (-1)·(1/4000)·g(y) = 0


( 1/ln( g(200) ) ) = 20€


m = 3 cajas & x = 60 cm ==> ( 1/ln( f(60) ) ) = 1200€


d_{y}[g(y)] + (-1)·(1/72000)·g(y) = 0


( 1/ln( g(1200) ) ) = 60€

economia: modelo de línea de la ecuación lineal de primer orden

f(x) = e^{(m/x)}
ln( f(x) ) = (m/x)


d_{y}[g(y)] + (-1)·( m/x^{2} )·g(y) = 0


g(y) = e^{(m/x^{2})·y}


(1/ln(e^{(m/x^{2})·(x/m)})) = x


(1/ln(e^{(m/x^{2})·( 1/ln( f(x) ) )})) = x


( 1/ln( g( 1/ln(f(x)) ) ) ) = x


m = 4 botellas & x = 64 cm ==> ( 1/ln( f(64) ) ) = 16€


d_{y}[g(y)] + (-1)·(1/1024)·g(y) = 0


( 1/ln( g(16) ) ) = 64€


m = 3 botellas & x = 30 cm ==> ( 1/ln( f(30) ) ) = 10€


d_{y}[g(y)] + (-1)·(1/300)·g(y) = 0


( 1/ln( g(10) ) ) = 30€

micro-economia optimizatció de una esfera

B(r) = (1/10pi)·(4pi)·( p·r^{2}+(-1)·m·(1/3)·r^{3} )


d_{x}[B(r)] = 2p·r+(-1)·m·r^{2}


mr = 2p


r = ( (2p)/m )


B( (2p)/m ) = (4pi)( (2/3)(p^{3}/m^{2}) )


pelota de tenis:


p=3 &  m=2


B(3) = (1/10pi)·(18pi) = 1.8€
B(r) = (1/10pi)·( F(r)+(-1)·V(r) )
(1/10pi)·F(3)  = 2.7€
(1/10pi)·V(3) = 0.9€

micro-economia optimització de una caisha

B(x,y,z) = (1/4000)·( 2·p·( xy+yz+zx )+(-1)·m·xyz )


d_{x}[B(x,y,z)] = 2·p·( y+z )+(-1)·m·yz
d_{y}[B(x,y,z)] = 2·p·( z+x )+(-1)·m·zx
d_{z}[B(x,y,z)] = 2·p·( x+y )+(-1)·m·xy


2·p·( y+z )=m·yz
2·p·( z+x )=m·zx
2·p·( x+y )=m·xy


( y+z )=m
( z+x )=m
( x+y )=m


2·p=yz
2·p=zx
2·p=xy


x=(m/2)
y=(m/2)
z=(m/2)


p=(1/2)·(m/2)^{2}


B( (m/2) , (m/2) , (m/2) ) = 3·(m/2)^{4}+(-1)·2·(m/2)^{4}
B( (m/2) , (m/2) , (m/2) ) = (m/2)^{4}


bombones
m=40 <==> ( x=20 & y=20 & z=20 )


B( 20 , 20 , 20 ) = 4€


B( x , y , z ) = (1/4000)·( F( x , y , z ) +(-1)·V( x , y , z )  )


(1/4000)·F( 20 , 20 , 20 ) = 12€
(1/4000)·V( 20 , 20 , 20 ) = 8€

micro-economia optimizar un rollo de papel

B(r,z) = (1/10pi)·( p·(2pi)·r·z+(-1)·m·(pi)·r^{2}·z )


d_{r}[B(r,z)]  = p·(2pi)·z+(-1)·m·(2pi)·r·z


p = m·r


r = (p/m)


B( (p/m) , z ) = (1/10)·( p·(2pi)·(p/m)·z+(-1)·m·(pi)·(p/m)^{2}·z )


B( (p/m) , z ) = (1/10)·( (p^{2}/m)·(pi)·z )


papel higiénico:
z=10 & p=(5céntimos/(cm)^{2}) & m=(1céntimo/(cm)^{3})


B( 5 , 10  ) = 25·céntimos = 0.25€


B(r,z) = (1/10pi)·( F(r,z)+(-1)·V(r,z) )


(1/10pi)·F( 5 , 10  ) = 50·céntimos = 0.50€
(1/10pi)·V( 5 , 10  ) = 25·céntimos = 0.25€


papel de aluminio:
z=30 & p=(6céntimos/(cm)^{2}) & m=(2céntimo/(cm)^{3})


B( 3 , 30  ) = 54·céntimos = 0.54€


B(r,z) = (1/10pi)·( F(r,z)+(-1)·V(r,z) )


(1/10pi)·F( 3 , 30  ) = 108·céntimos = 1.08€
(1/10pi)·V( 3 , 30  ) = 54·céntimos = 0.54€

economia: simetria-dual de consumo segun un capital producto lineal en no uno

F(x,y) = (n·x)·(m·y)+(-1)·h( px+qy+(-k) ) & px+qy=k


d_{x}[F(x,y)] = n(my)+(-1)·hp
d_{y}[F(x,y)] = m(nx)+(-1)·hq


nmy=hp
mnx=hq


nmyx=hpx
mnxy=hqy


nmyx+mnxy=hk


2·mnxy=hk


( x=( e^{n}/n ) & y=( e^{m}/m ) )  <==>  2·e^{n}e^{m}=hk


si h=2·( e^{k}/k ) ==>


e^{n}e^{m}=e^{k}


( n=( k+(-j) ) & m=j )


G( (e^{n}/n ) , (e^{m}/m ) ) = e^{n}·e^{m}

economia: simetria-dual de consumo segun un capital producto lineal

F(x,y) = (e^{n}·x)·(e^{m}·y)+(-1)·h( px+qy+(-k) ) & px+qy=k


d_{x}[F(x,y)] = e^{n}(e^{m}y)+(-1)·hp
d_{y}[F(x,y)] = e^{m}(e^{n}x)+(-1)·hq


e^{n}e^{m}y=hp
e^{m}e^{n}x=hq


e^{n}e^{m}yx=hpx
e^{m}e^{n}xy=hqy


e^{n}e^{m}yx+e^{m}e^{n}xy=hk


2·e^{n}e^{m}xy=hk


( x=1 & y=1 )  <==>  2·e^{n}e^{m}=hk


si h=2·( e^{k}/k ) ==>


e^{n}e^{m}=e^{k}


( n=( k+(-j) ) & m=j )


G( 1 , 1 ) = e^{n}·e^{m}

economia: simetria-dual de consumo segun un capital producto potencial

F(x,y) = x^{n}·y^{m}+(-1)·h( px+qy+(-k) ) & px+qy=k


d_{x}[F(x,y)] = nx^{(n+(-1))}·y^{m}+(-1)·hp
d_{y}[F(x,y)] = my^{(m+(-1))}·x^{n}+(-1)·hq


nx^{(n+(-1))}·y^{m}=hp
my^{(m+(-1))}·x^{n}=hq


nx^{n}·y^{m}=hpx
my^{m}·x^{n}=hqy


n·x^{n}·y^{m}+m·x^{n}·y^{m}=hk


(n+m)·x^{n}·y^{m}=hk


( x=1 & y=1 )  <==>  n+m=hk


si h=1 ==>


( n=( k+(-j) ) & m=j )


G( 1 , 1 ) = 1

economia: simetria-dual de consumo segun un capital exponencial

F(x,y) = e^{nx}+e^{my}+(-1)·h( px+qy+(-k) ) & px+qy=k


d_{x}[F(x,y)] = ne^{nx}+(-1)·hp
d_{y}[F(x,y)] = me^{my}+(-1)·hq


ne^{nx}=hp
me^{my}=hq


nxe^{nx}=hpx
mye^{my}=hqy


nxe^{nx}+mye^{my}=hk


x=( (k+(-j))/n )
y=( j/m )


(k+(-j))e^{k+(-j)}+je^{j}=hk


si h=( (k+(-j))e^{k+(-j)}+je^{j})/k ) ==>


( x=1 & y=1 ) <==> ( n=( k+(-j) ) & m=j )


G( 1 , 1 ) = e^{n}+e^{m}

economia: simetria-dual de consumo segun un capital logarítmica

F(x,y) = n·ln(x)+m·ln(y)+(-1)·h( px+qy+(-k) ) & px+qy=k


d_{x}[F(x,y)] = n·(1/x)+(-1)·hp
d_{y}[F(x,y)] = m·(1/y)+(-1)·hq


n·(1/x)=hp
m·(1/y)=hq


n=hpx
m=hqy


n+m=hk


si h=1 ==>


n=k+(-j)
m=j


G( n/p , m/q ) = n·ln(n/p)+m·ln(m/q)

economia simetria-dual de consumo segun un capital lineal

F(x,y) = nx+my+(-1)·h( px+qy+(-k) ) & px+qy = k


d_{x}[F(x)] = n+(-1)hp
d_{y}[F(x)] = m+(-1)hq


n=hp
m=hq


nx=hpx
my=hqy


nx+my=hk


si h=1 ==>


x=( (k+(-j))/n )
y=( j/m )


( x=1 & y=1 ) <==> ( n = ( k+(-j) ) & m = j )


G( 1 , 1 ) = n+m

economia simetria-dual de consumo segun un capital potencial

F(x,y) = x^{n}+y^{m}+(-1)·h( px+qy+(-k) ) & px+qy = k


d_{x}[F(x,y)] = nx^{n+(-1)}+(-1)·hp
d_{y}[F(x,y)] = my^{m+(-1)}+(-1)·hq


nx^{n+(-1)}=hp
my^{m+(-1)}=hq


nx^{n}=hpx
my^{m}=hqy


nx^{n}+my^{m}=hk


x=( ((k+(-j))h)/n )^{(1/n)}
y=( (jh)/m )^{(1/m)}


Si h=1 ==>


x=( (k+(-j))/n )^{(1/n)}
y=( j/m )^{(1/m)}


( x=1 & y=1 ) <==> ( n=( k+(-j) ) & m=j )


G( 1 , 1 ) = 2

micro-economia supermercado

un supermercado tiene unos costes en camión de alimentos de mx^{3},
optimizar el precio de los alimentos n si los compra a precio p:


B(x) = (n+(-p))·x+(-1)·mx^{3}


d_{x}[B(x)] = (n+(-p))+(-1)·(3mx^{2})


d_{x}[B(x)] = 0  <==> x = ( (n+(-p))/3m )^{(1/2)}


x=1<==> 3m=(n+(-p))


( m=10€ de gasolina & 200€ la planta & n=230€ la planta )
( m=15€ de gasolina & 200€ la planta & n=245€ la planta )
( m=20€ de gasolina & 200€ la planta & n=260€ la planta )


B(1) = (n+(-p))+(-1)·m
B(1) = 2·m

micro-economia de coste cuadrático

una plantación tiene unos costes en agua de qx^{2},
optimizar el precio de la planta:


B(x) = px+(-1)·qx^{2}


d_{x}[B(x)] = p+(-1)·(2qx)


d_{x}[B(x)] = 0  <==> x = ( p/2q )


x=1<==> 2q=p


( m=100€ de agua & p=200€ la planta )
( m=150€ de agua & p=300€ la planta )
( m=200€ de agua & p=400€ la planta )


B(1) = p+(-1)·q
B(1) = q

micro-economia de coste cúbico

una empresa de gas tiene unos costes en camión de bombonas de mx^{3},
optimizar el precio de la bombona de gas:


B(x) = px+(-1)·mx^{3}


d_{x}[B(x)] = p+(-1)·(3mx^{2})


d_{x}[B(x)] = 0  <==> x = ( p/3m )^{(1/2)}


x=1<==> 3m=p


( m=10€ de gasolina & p=30€ la bombona de gas )
( m=15€ de gasolina & p=45€ la bombona de gas )
( m=20€ de gasolina & p=60€ la bombona de gas )


B(1) = p+(-1)·m
B(1) = 2·m

micro-economia

f(x) = px <==> preu per unitat.
g(x) = qx^{2} <==> preu per superficie.
h(x) = mx^{3} <==> preu per volum.


d_{x}[f(x)] = p
d_{x}[g(x)] = 2qx
d_{x}[h(x)] = 3mx^{2}