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Solve using Lagrange multipliers. Find three positive numbers whose sum is 36 and whose product is as large as possible.

Short Answer

Expert verified
The numbers are 12, 12, and 12.

Step by step solution

01

Define the objective function

In this problem, we want to maximize the product of three positive numbers which we will call \( x \), \( y \), and \( z \). The objective function to maximize is \( f(x, y, z) = xyz \).
02

Define the constraint equation

We are given that the sum of the three numbers is 36. This gives us the constraint equation \( g(x, y, z) = x + y + z - 36 = 0 \).
03

Set up the Lagrange function

To use Lagrange multipliers, we set up the Lagrangian function \( \mathcal{L}(x, y, z, \lambda) = xyz + \lambda (36 - x - y - z) \), where \( \lambda \) is the Lagrange multiplier.
04

Compute partial derivatives

Find the partial derivatives of the Lagrangian with respect to \( x \), \( y \), \( z \), and \( \lambda \). They are:\[ \frac{\partial \mathcal{L}}{\partial x} = yz - \lambda = 0 \]\[ \frac{\partial \mathcal{L}}{\partial y} = xz - \lambda = 0 \]\[ \frac{\partial \mathcal{L}}{\partial z} = xy - \lambda = 0 \]\[ \frac{\partial \mathcal{L}}{\partial \lambda} = 36 - x - y - z = 0 \]
05

Set partial derivatives equal to zero

From the first three equations, we get:\[ yz = xz = xy = \lambda \]Equate the terms to get the relationships:\[ yz = xz \Rightarrow x = y \]\[ xz = xy \Rightarrow y = z \]\[ xy = yz \Rightarrow x = z \]So, \( x = y = z \).
06

Solve for variable values

From the constraint equation \( x + y + z = 36 \) and knowing \( x = y = z \), we substitute to get:\[ 3x = 36 \]Solving for \( x \), we find \( x = 12 \). Thus, \( y = 12 \) and \( z = 12 \).
07

Verify the solution

Finally, we check the original constraint and objective:- The sum: \( 12 + 12 + 12 = 36 \).- The product: \( 12 \cdot 12 \cdot 12 = 1728 \), which satisfies the condition of being the largest possible.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Optimization
In mathematics, optimization refers to the process of finding the best solution from a set of possible choices. This can mean maximizing or minimizing a particular function.
In the context of Lagrange multipliers, optimization involves finding the maximum (or minimum) value of an objective function while satisfying certain constraints. Here, we are looking at maximizing the product of three numbers.
The main goal is to determine the optimal values that provide the largest product under the limitation that their sum equals 36. Hence, optimization with constraints is a common practice in both mathematical problems and real-world applications.
Constraint Equation
A constraint equation explicitly defines the condition that must be satisfied in an optimization problem. It acts as a boundary for the system we aim to optimize. For the exercise at hand, the constraint is described by the equation:
- The sum of the three variables should equal 36. - Mathematically, this is expressed as: \[( x + y + z - 36 = 0 )\]
This constraint ensures that any values chosen for the numbers will adhere to this rule at all times. Constraints like these are often necessary to reflect real-life scenarios where resources are limited or specific requirements must be respected.
Partial Derivatives
Partial derivatives provide a way to understand how a function changes when one of several variables is varied while the others are kept constant.
In the Lagrangian method, partial derivatives are calculated for each variable involved to find critical points where a function is optimized.
Here, we find the partial derivatives of the Lagrangian function with respect to each variable, which reveals the necessary conditions for maximum or minimum values. - For example, \[ \frac{\partial \mathcal{L}}{\partial x} = yz - \lambda = 0 \] is one of these derivatives.
Solving these equations helps determine the values of the variables that maximize the product under the given constraints.
Objective Function
The objective function in an optimization problem is the function that needs to be maximized or minimized.
For our specific problem, the objective function is the product of the three positive numbers. This function is expressed as: - \[ f(x, y, z) = xyz \] It represents what we are aiming to optimize, which in this case is to make the product as large as possible.
The effectiveness of the optimization process depends on correctly identifying and formulating this objective, as it directly dictates how other methods, like using Lagrange multipliers, work to find optimal solutions.

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