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

Short Answer

Expert verified
The largest value of the product is \(\frac{1}{27}\).

Step by step solution

01

- Define the Objective Function

Let the three positive numbers be denoted as \(x, y,\) and \(z\). The objective is to maximize their product, so define the objective function as \(f(x, y, z) = xyz\).
02

- Define the Constraint

The given constraint is that the sum of the three numbers is 1. Thus, the constraint function is \(g(x, y, z) = x + y + z - 1 = 0\).
03

- Formulate the Lagrangian

Construct the Lagrangian function \(L(x, y, z, \lambda) = xyz + \lambda(1 - x - y - z)\), where \(\lambda\) is the Lagrange multiplier.
04

- Compute the Partial Derivatives

Calculate the partial derivatives of \(L\) with respect to \(x, y, z,\) and \(\lambda\), and set them equal to zero:\(\frac{\partial L}{\partial x} = yz - \lambda = 0\)\(\frac{\partial L}{\partial y} = xz - \lambda = 0\)\(\frac{\partial L}{\partial z} = xy - \lambda = 0\)\(\frac{\partial L}{\partial \lambda} = 1 - x - y - z = 0\)
05

- Solve the System of Equations

From the equations \(yz = \lambda\), \(xz = \lambda\), and \(xy = \lambda\), it follows that \(yz = xz = xy\). Consequently, we find that \(x = y = z\). Substituting \(x = y = z\) into the constraint equation \(x + y + z = 1\), we get \(3x = 1\) or \(x = \frac{1}{3}\). Thus, \(y = \frac{1}{3}\) and \(z = \frac{1}{3}\).
06

- Evaluate the Objective Function

Substitute \(x = \frac{1}{3}\), \(y = \frac{1}{3}\), and \(z = \frac{1}{3}\) into the objective function \(f(x, y, z) = xyz\):\(f\left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right) = \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{3} = \frac{1}{27}\).

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Key Concepts

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

Optimization
Optimization involves finding the best solution to a problem given certain constraints. In this exercise, we aim to maximize the product of three positive numbers. The key is to define an objective function that represents the quantity we want to optimize. Here, our objective function is the product of the three numbers: \(f(x, y, z) = xyz\). The objective function tells us what we are optimizing. The optimization process requires leveraging mathematical techniques to find the values of the variables that maximize (or minimize) this function, given any constraints.
Constraint Equations
Constraint equations represent the limitations or requirements that must be satisfied. In this exercise, our constraint is that the sum of the three numbers equals 1: \(x + y + z = 1\). Constraints are critical because they limit the solution space of the problem. Without the constraint, we could choose any values for \(x, y,\) and \(z\), but with it, only specific combinations that satisfy the constraint are feasible. To incorporate constraints into the optimization, we use the method of Lagrange multipliers, which introduces a new variable (the Lagrange multiplier) to account for the constraint.
Partial Derivatives
Partial derivatives are used to find the rate of change of a function with respect to one of its variables while keeping the other variables constant. In the Lagrange multiplier method, we take the partial derivatives of the Lagrangian function with respect to each variable. The Lagrangian function combines the objective function and the constraint: \(L(x, y, z, \lambda) = xyz + \lambda(1 - x - y - z)\). To find the optimal values, set the partial derivatives equal to zero:
  • \(\frac{\partial L}{\partial x} = yz - \lambda = 0\)
  • \(\frac{\partial L}{\partial y} = xz - \lambda = 0\)
  • \(\frac{\partial L}{\partial z} = xy - \lambda = 0\)
  • \(\frac{\partial L}{\partial \lambda} = 1 - x - y - z = 0\)
Setting these equations to zero and solving them helps us identify the critical points where the function may achieve its maximum or minimum subject to the given constraints.
Maximum Product
The goal of this exercise is to find the maximum product of three positive numbers whose sum is 1. After setting up the system of equations from the partial derivatives, we find that \(x = y = z\). Substituting into the constraint \(x + y + z = 1\), we get \(3x = 1\), hence \(x = \frac{1}{3}\). Thus, \(y = \frac{1}{3}\) and \(z = \frac{1}{3}\). Finally, substituting these values into the objective function gives: \(f(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}) = \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{3} = \frac{1}{27}.\) Therefore, the largest value of the product of three positive numbers whose sum is 1 is \(\frac{1}{27}.\)

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Most popular questions from this chapter

If \(u=x^{2}+y^{2}+x y z\) and \(x^{4}+y^{4}+z^{4}=2 x^{2} y^{2} z^{2}+10,\) find \((\partial u / \partial x)_{z}\) at the point \((x, y, z)=(2,1,1)\).

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