Question

Find by the Lagrange multiplier method the largest value of the product of three positive numbers if their sum is 1 .

Short answer

The maximum product is \( \frac{1}{27} \).

Step-by-step solution

Define the objective function

Let the three positive numbers be denoted as \(x\), \(y\), and \(z\). The objective is to maximize the product \(P = xyz\).

Define the constraint

The given constraint is that the sum of the three numbers is 1, which can be written as \(x + y + z = 1\).

Set up the Lagrange function

Construct the Lagrange function \( \mathcal{L}(x, y, z, \lambda) = xyz - \lambda (x + y + z - 1) \), where \(\lambda\) is the Lagrange multiplier.

Take partial derivatives

Compute the partial derivatives of \( \mathcal{L} \) with respect to \(x\), \(y\), \(z\), and \(\lambda\), and set them equal to zero:1. \( \frac{\partial \mathcal{L}}{\partial x} = yz - \lambda = 0 \)2. \( \frac{\partial \mathcal{L}}{\partial y} = xz - \lambda = 0 \)3. \( \frac{\partial \mathcal{L}}{\partial z} = xy - \lambda = 0 \)4. \( \frac{\partial \mathcal{L}}{\partial \lambda} = -(x + y + z - 1) = 0 \)

Solve the system of equations

From the partial derivatives with respect to \(x\), \(y\), and \(z\), we get:1. \( yz = \lambda \)2. \( xz = \lambda \)3. \( xy = \lambda \)From these equations, we observe that \( yz = xz = xy = \lambda\), therefore, \(x = y = z\).

Substitute the constraint

Substitute \( x = y = z \) into the constraint equation \( x + y + z = 1 \). So \( 3x = 1 \), hence \( x = \frac{1}{3} \). Therefore, \( y = \frac{1}{3} \), and \( z = \frac{1}{3} \).

Calculate the product

The product is then \( P = xyz = \left( \frac{1}{3} \right) \left( \frac{1}{3} \right) \left( \frac{1}{3} \right) = \frac{1}{27} \).

Key concepts

Optimization

Optimization is all about finding the best possible solution to a problem. In mathematics, especially in calculus, we often aim to maximize or minimize a function. For example, we can maximize the product of three numbers given that their sum equals a specific value. To solve these problems, we need to use different techniques.
Lagrange multipliers is one such technique. It is particularly useful for optimization problems with constraints. This means we want to maximize or minimize a function but have some conditions that the solution must satisfy.

Multivariable Calculus

Multivariable calculus extends the concepts of calculus to functions of several variables. Unlike single-variable calculus, where you deal with functions of one variable, here we handle functions involving two or more variables.
For example, if we have a function of three variables like \(f(x, y, z)\), we can analyze how this function behaves when the variables change.
Multivariable calculus is crucial for solving optimization problems, especially when constraints are involved.
Using Lagrange multipliers, we can handle complex functions and constraints effectively.

Constraint Equations

A constraint equation defines the conditions that must be satisfied for a solution to be valid. In optimization problems, constraints limit the possible solutions.
For instance, if we have three positive numbers whose sum must be 1, we have a constraint equation \(x + y + z = 1\).
To find the maximum product of these numbers, we need to consider this constraint. This is where Lagrange multipliers become useful. By incorporating the constraint directly into the objective function, we can solve for the maximum or minimum values.

Partial Derivatives

In multivariable calculus, partial derivatives are used to analyze how a function changes as one of its variables changes, while the others are held constant.
When applying the Lagrange multipliers method, we take partial derivatives of the Lagrange function with respect to each variable.
For example, if our Lagrange function is \(\text{L}(x, y, z, abla) = xyz - abla (x + y + z - 1)\), we take partial derivatives with respect to \(x\), \(y\), \(z\), and the Lagrange multiplier \(abla\).
Setting these partial derivatives equal to zero helps us find the critical points that satisfy both the objective function and the constraints.