Nonlinear constraint optimization in SciPy - Mini Project: Build & Apply

Nonlinear Constraint Optimization with SciPy

📖 Scenario: Imagine you are helping a small business owner who wants to maximize their profit by deciding how many units of two products to produce. However, there are limits on resources and production rules that must be followed.

🎯 Goal: You will build a program that finds the best number of units for each product to maximize profit while respecting the business constraints using nonlinear constraint optimization.

📋 What You'll Learn

Create a function representing the profit to maximize

Define nonlinear constraints for the problem

Use SciPy's minimize function with method 'SLSQP' to solve the problem

Print the optimal production quantities and maximum profit

💡 Why This Matters

🌍 Real World

Businesses often need to maximize profits or minimize costs while respecting limits on resources, labor, or materials. Nonlinear constraints model real-world rules that are not simple linear limits.

💼 Career

Understanding nonlinear constraint optimization is valuable for roles in operations research, data science, and analytics where decision-making under constraints is common.

1

Define the profit function

Create a function called profit that takes a list x with two elements representing the units of product A and product B. The profit is calculated as 40 * x[0] + 30 * x[1]. Return the negative of this value because SciPy minimizes functions.

SciPy

# Define the profit function that takes x and returns negative profit
# Your code here

Need a hint?

Remember, SciPy's minimize finds minimum values, so return negative profit to maximize.

2

Set up nonlinear constraints

Create a list called constraints with two nonlinear constraints as dictionaries. The first constraint function constraint1 ensures 2 * x[0] + x[1] <= 100. The second constraint function constraint2 ensures x[0] + 2 * x[1] <= 80. Use type: 'ineq' for both constraints, meaning the function should be >= 0.

SciPy

def profit(x):
    return -(40 * x[0] + 30 * x[1])

# Define constraint1 function
# Your code here

Need a hint?

Constraints must be functions returning values >= 0 for valid solutions.

3

Solve the optimization problem

Use SciPy's minimize function to find the optimal units of products A and B. Use initial guess x0 = [0, 0], method 'SLSQP', and the constraints list. Store the result in a variable called result.

SciPy

def profit(x):
    return -(40 * x[0] + 30 * x[1])

def constraint1(x):
    return 100 - (2 * x[0] + x[1])

def constraint2(x):
    return 80 - (x[0] + 2 * x[1])

constraints = [
    {'type': 'ineq', 'fun': constraint1},
    {'type': 'ineq', 'fun': constraint2}
]

# Import minimize from scipy.optimize
# Use minimize with profit, x0=[0, 0], method='SLSQP', and constraints
# Store result in variable result
# Your code here

Need a hint?

Remember to import minimize from scipy.optimize before using it.

4

Print the optimal solution and profit

Print the optimal units of product A and product B from result.x with labels. Then print the maximum profit by negating result.fun.

SciPy

from scipy.optimize import minimize

def profit(x):
    return -(40 * x[0] + 30 * x[1])

def constraint1(x):
    return 100 - (2 * x[0] + x[1])

def constraint2(x):
    return 80 - (x[0] + 2 * x[1])

constraints = [
    {'type': 'ineq', 'fun': constraint1},
    {'type': 'ineq', 'fun': constraint2}
]

result = minimize(profit, x0=[0, 0], method='SLSQP', constraints=constraints)

# Print the optimal units for product A and B
# Print the maximum profit
# Your code here

Need a hint?

Use result.x for units and -result.fun for profit. Format numbers to two decimals.