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€