Question

  A company makes four types of gourmet chocolate bars. Chocolate Bar I contains 5 grams of nuts, 5 grams of biscuit, 10 grams of caramel, and 100 grams of chocolate, and sells for $5.40. Chocolate Bar II contains 10 grams of nuts, 10 grams of biscuit, 10 grams of caramel, and 90 grams of chocolate and sells for $6.25. Chocolate Bar III contains no nuts, 10 grams of biscuit, 10 grams of caramel, and 100 grams of chocolate and sells for $5.25. Chocolate Bar IV contains 20 grams of nuts, no biscuit, 10 grams of caramel, and 90 grams of chocolate and sells for $7.00. The ingredient costs are $0.15 per gram of nuts, $0.025 per gram of biscuit, $0.02 per gram of caramel, and $0.015 per gram of chocolate. The company has supplier agreements that force them to order at least 50,000 grams of each ingredient per week (nuts, biscuit, caramel, and chocolate). However, due to the space in their warehouse and the expense of holding too much inventory, the company cannot hold more than 100,000 grams of nuts; 75,000 grams of biscuits; 85,000 grams of caramel; and 500,000 grams of chocolate per week. How many chocolate bars of each type should the company produce each week in order to maximize its weekly profit?   

Answer 1

Let be the number of chocolate bars of type I, II, III, and IV respectively.

Amount of nuts needed

Amount of biscuits needed

Amount of caramel needed

Amount of chocolate needed

Therefore, cost price of nuts

Cost price of biscuits

Cost price of caramel

Cost price of chocolate

Total cost

Total sale

Therefore, profit

We want to maximize the objective function

subject to

You will get maximum at and the maximum profit

Write a Python code as follows:

import numpy as np
from scipy.optimize import minimize

def objective(x):
return x[0]*x[3]*(x[0]+x[1]+x[2])+x[2]

def constraint1(x):
return x[1]
def constraint2(x):
return x[2]
def constraint3(x):
return x[3]
def constraint4(x):
return x[4]
def constraint5(x):
return x[1]+2*x[2]+4*x[4]-10000
def constraint6(x):
return 20000-(x[1]+2*x[2]+4*x[4])
def constraint7(x):
return x[1]+2*x[2]+2*x[3]-10000
def constraint8(x):
return 15000-(x[1]+2*x[2]+2*x[3])
def constraint9(x):
return x[1]+x[2]+x[3]+x[4]-5000
def constraint10(x):
return 8500-(x[1]+x[2]+x[3]+x[4])
def constraint11(x):
return 10*x[1]+9*x[2]+10*x[3]+9*x[4]-5000
def constraint12(x):
return 50000-(10*x[1]+9*x[2]+10*x[3]+9*x[4])
# initial guesses
n = 4
x0 = np.zeros(n)
x0[0] = 0.0
x0[1] = 0.0
x0[2] = 0.0
x0[3] = 0.0

# show initial objective
print('Initial SSE Objective: ' + str(objective(x0)))

# optimize
b = (0.0,8500.0)
bnds = (b, b, b, b)
con1 = {'type': 'ineq', 'fun': constraint1}
con2 = {'type': 'ineq', 'fun': constraint2}
con3 = {'type': 'ineq', 'fun': constraint3}
con4 = {'type': 'ineq', 'fun': constraint4}
con5 = {'type': 'ineq', 'fun': constraint5}
con6 = {'type': 'ineq', 'fun': constraint6}
con7 = {'type': 'ineq', 'fun': constraint7}
con8 = {'type': 'ineq', 'fun': constraint8}
con9 = {'type': 'ineq', 'fun': constraint9}
con10 = {'type': 'ineq', 'fun': constraint10}
con11 = {'type': 'ineq', 'fun': constraint11}
con12 = {'type': 'ineq', 'fun': constraint12}
cons = ([con1,con2,con3,con4,con5,con6,con7,con8,con9,con10,con11,con12])
solution = minimize(objective,x0,method='SLSQP',\
bounds=bnds,constraints=cons)
x = solution.x

# show final objective
print('Final SSE Objective: ' + str(objective(x)))

# print solution
print('Solution')
print('x1 = ' + str(x[0]))
print('x2 = ' + str(x[1]))
print('x3 = ' + str(x[2]))
print('x4 = ' + str(x[3]))