Problem 30 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let \(w=x_{1} x_{2} \cdots x_{n}\). (a) Maximize \(w\) subject to \(x_{1}+x_{2}+\cdots+x_{n}=1\) and all \(x_{i}>0 .\) (b) Use part (a) to deduce the famous Geometric MeanArithmetic Mean Inequality for positive numbers \(a_{1}, a_{2}, \ldots, a_{n} ;\) that is, $$ \sqrt[n]{a_{1} a_{2} \cdots a_{n}} \leq \frac{a_{1}+a_{2}+\cdots+a_{n}}{n} $$

Step-by-Step Solution

Answer

The maximum of \( w \) is achieved when \( x_i = \frac{1}{n} \) for all \( i \). This gives the AM-GM inequality.

Step 1: Setting Up the Problem

We aim to maximize the product \( w = x_1 x_2 \cdots x_n \) given the constraint \( x_1 + x_2 + \cdots + x_n = 1 \) with \( x_i > 0 \). This setup suggests using the method of Lagrange multipliers.

Step 2: Lagrange Function

Define the Lagrange function as \( \mathcal{L}(x_1, x_2, \ldots, x_n, \lambda) = x_1 x_2 \cdots x_n + \lambda(1 - x_1 - x_2 - \cdots - x_n) \).

Step 3: Partial Derivatives

Compute the partial derivatives: \( \frac{\partial \mathcal{L}}{\partial x_i} = x_1 x_2 \cdots x_n \frac{1}{x_i} - \lambda = 0 \) for each \( i \). Also, \( \frac{\partial \mathcal{L}}{\partial \lambda} = 1 - x_1 - x_2 - \cdots - x_n = 0 \).

Step 4: Finding Stationary Points

From \( \frac{\partial \mathcal{L}}{\partial x_i} = 0 \), we get \( x_i = x_j \) for all \( i, j \). Thus, \( x_1 = x_2 = \cdots = x_n = \frac{1}{n} \) to satisfy the constraint \( x_1 + x_2 + \cdots + x_n = 1 \).

Step 5: Calculate Maximum Product

Substitute \( x_i = \frac{1}{n} \) into \( w = x_1 x_2 \cdots x_n \) to get \( w = \left(\frac{1}{n}\right)^n \). So, the maximum value of \( w \) is \( \frac{1}{n^n} \).

Step 6: Deduction of AM-GM Inequality

To prove the AM-GM inequality for any set of positive numbers \( a_1, a_2, \ldots, a_n \), consider \( x_i = \frac{a_i}{b} \), where \( b = \frac{a_1 + a_2 + \cdots + a_n}{n} \). Applying the previous result, we have \((\frac{a_1}{b} \cdot \frac{a_2}{b} \cdots \frac{a_n}{b})^{1/n} \leq \frac{1}{n}\), leading to the AM-GM inequality \( \frac{a_1 a_2 \cdots a_n}{b^n} \leq 1 \), or equivalently \( \sqrt[n]{a_1 a_2 \cdots a_n} \leq \frac{a_1 + a_2 + \cdots + a_n}{n} \).

Key Concepts

Lagrange multipliers

Lagrange multipliers is a powerful method used in calculus for finding the local maxima and minima of a function subject to equality constraints. In our scenario, we want to maximize the product \( w = x_1 x_2 \cdots x_n \) while ensuring that the sum \( x_1 + x_2 + \cdots + x_n = 1 \) is satisfied.
This is achieved by introducing a new variable, \( \lambda \), known as a Lagrange multiplier. The Lagrange multiplier helps us take into account the constraint while seeking the points where \( w \) might reach a maximum or minimum.

• Define a new function, called the Lagrangian:
  \( \mathcal{L}(x_1, x_2, \ldots, x_n, \lambda) = x_1 x_2 \cdots x_n + \lambda(1 - x_1 - x_2 - \cdots - x_n) \)
• Take partial derivatives of this function with respect to each variable and \( \lambda \).
• Set these partial derivatives to zero to find the stationary points, leading us to the conditions that maximize or minimize our function.

This approach transforms constraint problems into unconstrained ones, simplifying the search for solutions.

maximum product problem

The problem we are tackling is a classic example of the maximum product problem, which falls under the kind of problems where we try to maximize the product of variables subject to certain conditions. Here, we're tasked with maximizing \( w = x_1 x_2 \cdots x_n \) under the constraint that their sum is 1, and each variable is positive.
To do this effectively, we first set up our initial problem constraint and employ derivatives to explore its critical points. By fully analyzing these points, we can identify the configuration of \( x_i's \) that yields the highest possible product value.

• Using the result where the critical points showed equality (i.e., \( x_1 = x_2 = \cdots = x_n \)), the maximal configuration was found to be when each \( x_i = \frac{1}{n} \).
• This arrangement balances the distribution of values across the variables to maximize their product under the given sum constraint.

The solution shows that symmetrical distribution often helps achieve maximum product results in such scenarios.

constraint optimization

Constraint optimization involves optimizing a function while adhering to specific constraints. In our exercise, we work with the function \( w = x_1 x_2 \cdots x_n \) and the constraint \( x_1 + x_2 + \cdots + x_n = 1 \). The goal here is to find the values of \( x_1, x_2, \ldots, x_n \) that maximize \( w \) while not violating the constraint.
The Lagrange multipliers method is instrumental in handling such problems since it directly incorporates the constraints into the optimization process.

• First, we create a Lagrangian that marries our function and constraint.
• Next, we determine the partial derivatives with respect to each variable and the Lagrange multiplier.
• Lastly, solving the equations resulting from setting these derivatives to zero allows us to locate the optimal values for our original problem.

This process underscores the interdependence of calculus and algebra in solving optimization problems with constraints.

positive numbers

In our exercise, each \( x_i \) is conditioned to be a positive number. This constraint is essential for the validity of the problem, ensuring all components of the product \( w = x_1 x_2 \cdots x_n \) contribute meaningfully.
A positive number is simply any number greater than zero. Placing this condition means that all variables affect the product directly, rather than potentially nullifying it or flipping its sign, which wouldn't be meaningful in the context of this geometric mean-arithmetic mean inequality exercise.

• Maintaining positivity ensures that we are dealing with a maximization that remains strictly above zero.
• It also aligns with real-world scenarios where the quantities described cannot be negative, such as distances, areas, or probabilities.

Thus, the positivity requirement is crucial for ensuring the mathematical integrity and physical relevance of the solution.

mathematical inequality proof

The mathematical inequality proof achieved through this exercise is the famous Geometric Mean-Arithmetic Mean (GM-AM) Inequality. This inequality states that for any set of positive numbers \( a_1, a_2, \ldots, a_n \), the geometric mean is always less than or equal to the arithmetic mean:
\[ \sqrt[n]{a_1 a_2 \cdots a_n} \leq \frac{a_1 + a_2 + \cdots + a_n}{n} \]
By proving this with constraint optimization and Lagrange multipliers, we demonstrate how maximizing the product subject to a sum constraint naturally leads to this inequality.

• The arrangement \( x_1 = x_2 = \cdots = x_n \) which maximizes the product illustrates the point of equality in the inequality.
• This formulation reaffirms the intuitive balance achieved by equal distribution of variables under constraints.

The GM-AM inequality is a fundamental result with widespread applications in mathematics, science, and engineering.