Question

The number of grams of your favorite ice cream can be modeled by $$ G(x, y, z)=0.05 x^{2}+0.16 x y+0.25 z^{2} $$ where \(x\) is the number of fat grams, \(y\) is the number of carbohydrate grams, and \(z\) is the number of protein grams. Use Lagrange multipliers to find the maximum number of grams of ice cream you can eat without consuming more than 400 calories. Assume that there are 9 calories per fat gram, 4 calories per carbohydrate gram, and 4 calories per protein gram.

Short answer

The maximum grams of ice cream that can be consumed without exceeding 400 calories is obtained by solving the system of equations from the partial derivatives of the Lagrange function. The exact gram amount depends on the solution of the system.

Step-by-step solution

Setup the Lagrange Function

First set up the Lagrange function \( L \) that combines the gram function \( G \) and the constraint (calorie count). As per the problem statement, the total number of calories from fat, carb, and protein cannot exceed 400. This yields the equation \( 9x + 4y + 4z - 400 = 0 \) which is the constraint to be used. Now setup the Lagrange function \( L = G(x, y, z) - \lambda (9x + 4y + 4z - 400) \) where \( \lambda \) (lambda) is the Lagrange multiplier.

Find the Partial Derivatives

Lagrange multipliers method requires finding the points where the gradient of \( G \) is parallel to the gradient of the constraint. This is done by setting the partial derivatives of \( L \) with respect to \( x, y, z, \) and \( \lambda \) equal to zero and solving the resulting system of equations. So, compute the partial derivatives \( \frac{\partial L}{\partial x} = 0, \frac{\partial L}{\partial y} = 0, \frac{\partial L}{\partial z} = 0 \) and \( \frac{\partial L}{\partial \lambda} = 0 \)

Solve the Equations

After computing the derivatives, we have 4 equations. Solve this system of equations to obtain the values of \( x, y, z, \) and \( \lambda \).

Validation and Interpretation

Once the solution to the system of equations has been found, check if all numbers are non-negative as they represent grams. Then substitute these values back into \( G(x, y, z) \) to compute the maximum grams of ice cream you can eat.

Key concepts

Optimization

Optimization is the mathematical process of finding the best possible solution under given constraints and conditions. In this exercise, we are trying to maximize the grams of ice cream you can consume based on certain nutritional constraints.

For instance, the function \( G(x, y, z) = 0.05x^2 + 0.16xy + 0.25z^2 \) represents the grams of ice cream and depends on the amount of fat, carbohydrates, and protein. The goal is to find the maximum value of this function without exceeding the calorie limit.

Optimization problems like this often require finding a balance between different contributing factors. Here, calories from each nutrient play a crucial role, and we aim to maximize the grams of ice cream while staying within the caloric boundary of 400 calories.

Using mathematical tools like Lagrange multipliers helps streamline the process of balancing these aspects and finding that optimal solution.

Partial Derivatives

Partial derivatives are a critical concept when dealing with functions of multiple variables, especially in optimization problems. They represent the rate of change of a function concerning one variable while keeping the other variables constant.

In the context of the exercise, computing partial derivatives helps us understand how changes in fat, carbs, and protein affect the overall grams of ice cream, \( G(x, y, z) \).

By applying partial derivatives to the Lagrangian function, we set up a system of equations. Specifically, we compute \( \frac{\partial L}{\partial x} = 0 \), \( \frac{\partial L}{\partial y} = 0 \), \( \frac{\partial L}{\partial z} = 0 \), and \( \frac{\partial L}{\partial \lambda} = 0 \). Solving these equations helps us find the points where the gradient of the ice cream function is parallel to the constraint, leading us toward the optimal solution.

Understanding partial derivatives allows us to navigate the solution landscape efficiently, determining how individual nutrient changes impact the overall outcome.

Constraints in Calculus

Constraints in calculus play a vital role when optimizing functions, especially those with multiple variables. They represent restrictions or conditions that the solution must satisfy.

In this problem, the constraint is derived from the caloric content of the nutrients: \( 9x + 4y + 4z - 400 = 0 \). This equation ensures that the nutrient combinations must provide no more than 400 calories. The constraint is taken into account by incorporating it into the Lagrangian method, ensuring that the optimization process respects the calorie limit.

Constraints can be thought of as boundaries within which the solution must exist. By systematically including these conditions, we get a realistic and applicable solution to the problem at hand. Each constraint shapes the possible solutions, narrowing down the options to those that remain feasible.

Understanding how to incorporate these constraints is essential for using calculus effectively to tackle real-world problems involving optimization. In this manner, constraints guide us toward the most practical solution.