# Ejercicio 2.4 de la guia
#Maximizar z= 1000.A + 1500.B + 1800.C sujeto a:

# B1 + B2 = B

# Restricciones de capacidad   
# 5.A + 6.B1 <= 80
# 4.C + 4.B2 <= 80

#Restricicones de materia prima
# 1,6.A + 1,2.C <= 20
# 1,8.B1 + 1.8.B2 <= 56
# B1 + B2 >=10

# (A,B1,B2,C) e R > o

library(linprog)

Z1= c(1000,1500,1500,1800)

restric= c(5,0,1.6,0,0,6,0,0,1.8,1,0,4,0,1.8,1,0,4,1.2,0,0)
M= matrix(restric,ncol=4)
print(M)

b1=c(80,80,20,56,10)

inec= c("<=","<=","<=","<=",">=")

solucion.op= solveLP(Z1,b1,M,maximum = T, inec)

summary(solucion.op)