Problem 31

Question

Maximize $w=a_{1} x_{1}+a_{2} x_{2}+\cdots+a_{n} x_{n}$, all $a_{i}>0$, subject to $x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=1$.

Step-by-Step Solution

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Answer

The maximum value of $ w $ is obtained when each $ x_i = \frac{a_i}{2 \lambda} $ and they satisfy the constraint equation, where $ \lambda $ is found using $ \sum_{i=1}^n (\frac{a_i}{2 \lambda})^2 = 1 $.

1Step 1: Understand the Problem

We need to maximize the function $ w = a_1 x_1 + a_2 x_2 + \cdots + a_n x_n $, where all coefficients $ a_i $ are positive, subject to the constraint $ x_1^2 + x_2^2 + \cdots + x_n^2 = 1 $. This is a constrained optimization problem.

2Step 2: Set Up Lagrangian

Introduce a Lagrange multiplier $ \lambda $ and construct the Lagrangian $ \mathcal{L} = a_1 x_1 + a_2 x_2 + \cdots + a_n x_n + \lambda (1 - (x_1^2 + x_2^2 + \cdots + x_n^2)) $.

3Step 3: Compute Partial Derivatives

Differentiate $ \mathcal{L} $ with respect to each variable $ x_i $ and the Lagrange multiplier $ \lambda $. The derivatives with respect to $ x_i $ are $ \frac{\partial \mathcal{L}}{\partial x_i} = a_i - 2 \lambda x_i = 0 $, and with respect to $ \lambda $ is $ - (x_1^2 + x_2^2 + \cdots + x_n^2 - 1) = 0 $.

4Step 4: Solve the System of Equations

From $ \frac{\partial \mathcal{L}}{\partial x_i} = 0 $, we get $ x_i = \frac{a_i}{2 \lambda} $. Substitute these into the constraint $ x_1^2 + x_2^2 + \cdots + x_n^2 = 1 $ to find $ \lambda $: \[ \sum_{i=1}^n \left(\frac{a_i}{2 \lambda}\right)^2 = 1. \] Solve this equation to find $ \lambda $.

5Step 5: Find Maximum Value

Using the value of $ \lambda $ from Step 4, calculate each $ x_i $ using $ x_i = \frac{a_i}{2 \lambda} $. Substitute these $ x_i $ into $ w = a_1 x_1 + a_2 x_2 + \cdots + a_n x_n $ to find the maximum value of $ w $.

6Step 6: Verify the Solution

Ensure that the values of $ x_i $ satisfy the original constraint $ x_1^2 + x_2^2 + \cdots + x_n^2 = 1 $. Check for any calculation errors and confirm that the solution makes intuitive sense given the problem's constraints and the positive coefficients $ a_i $.

Key Concepts

Constrained OptimizationPartial DerivativesMaximum Value Solution

Constrained Optimization

In our exercise, we are dealing with a fascinating branch of mathematics called constrained optimization. This is a method used to find the best solution for a given function, given some constraints. Here, we want to maximize the function:

$ w = a_1 x_1 + a_2 x_2 + \cdots + a_n x_n $

All coefficients $a_i$ are positive numbers. But there's a catch: we need to ensure that the sum of the squares of the variables remains equal to one:

$ x_1^2 + x_2^2 + \cdots + x_n^2 = 1 $

This is the constraint we have to consider.
Think of constrained optimization as trying to find the best path under specific rules. It helps us tackle real-world challenges, where maximizing or minimizing outcomes while adhering to certain boundaries is crucial. To solve such problems, mathematicians often use a technique called Lagrange multipliers, which introduces a new variable to incorporate the constraint into the overall problem setup.

Partial Derivatives

Partial derivatives have a vital role in our optimization problem. When working with the Lagrangian, we define the Lagrangian for our problem by introducing a multiplier $ \lambda $:

$ \mathcal{L} = a_1 x_1 + a_2 x_2 + \cdots + a_n x_n + \lambda (1 - (x_1^2 + x_2^2 + \cdots + x_n^2)) $

Partial derivatives are used to find how the function $ \mathcal{L} $ changes when we vary the variables $x_i$ and the multiplier $\lambda$.
This gives us the following derivatives:

$ \frac{\partial \mathcal{L}}{\partial x_i} = a_i - 2 \lambda x_i = 0 $
With respect to $\lambda$: $ (x_1^2 + x_2^2 + \cdots + x_n^2 - 1) = 0 $

By solving these equations, we can express $x_i$ in terms of $\lambda$. These equations help us adjust the variables until we find the best result where the constrained function is optimized.
Partial derivatives are like tools that let you peek into the heart of a function, seeing precisely how it shifts and bends with tiny changes to each variable.

Maximum Value Solution

The crux of our exercise is to reach the maximum value of $ w $. After finding expressions for $ x_i $ in terms of the Lagrange multiplier $ \lambda $,

$ x_i = \frac{a_i}{2 \lambda} $

We substitute these back into the constraint,

$ \sum_{i=1}^n \left(\frac{a_i}{2 \lambda}\right)^2 = 1 $

This allows us to determine $ \lambda $. Once $ \lambda $ is known, we can plug it back to find each $ x_i $ and finally, calculate the maximum value of the original function $ w $.
In this way, Lagrange multipliers help us solve the puzzle of constrained optimization. They guide us to the top of the mountain,

Problem 30

Problem 31

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