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Solve the optimization problems. Minimize \(S=x+2 y\) with \(x y=2\) and both \(x\) and \(y>0\).

Short Answer

Expert verified
The minimum value of the function S(x, y) = x + 2y with the constraint xy = 2 and both x and y > 0 is S(2, 1) = 4.

Step by step solution

01

Define the given function and constraint

We have the function to minimize, S(x, y) = x + 2y, and the constraint G(x, y) = xy - 2 = 0.
02

Set up the Lagrange function

We will set up the Lagrange function L(x, y, λ) by adding the constraint multiplied by a Lagrange multiplier (λ) to the given function. So, L(x, y, λ) = S(x, y) + λG(x, y) = x + 2y + λ(xy - 2).
03

Calculate partial derivatives

Now, we need to find the partial derivatives of L(x, y, λ) with respect to x, y, and λ and set them equal to zero to solve for the optimal x, y, and λ. So, we have: \(\frac{\partial L}{\partial x} = 1 + yλ = 0\) \(\frac{\partial L}{\partial y} = 2 + xλ = 0\) \(\frac{\partial L}{\partial λ} = xy - 2 = 0\)
04

Solve the system of equations

We have three equations: 1. \(1 + yλ = 0\) 2. \(2 + xλ = 0\) 3. \(xy - 2 = 0\) From equation 1, we have \(λ = -\frac{1}{y}\), and from equation 2, we have \(λ = -\frac{2}{x}\). Setting them equal, we get: \(-\frac{1}{y} = -\frac{2}{x} \implies x = 2y\) Now we can substitute the value of x back to the constraint (equation 3): \((2y)y - 2 = 0 \implies 2y^2 - 2 = 0\) Divide both sides by 2 and solve for y: \(y^2 = 1 \implies y = 1\) (since both x and y are greater than 0) Substitute the value of y back to the relation between x and y found earlier, \(x = 2y \implies x = 2(1) \implies x = 2\)
05

Find the minimum value

Now that we have the optimal values of x and y, we can find the minimum value of S(x, y) using these values: S(x, y) = x + 2y = 2 + 2(1) = 4 So, the minimum value of the function S(x, y) = x + 2y with the constraint xy = 2 and both x and y > 0 is S(2, 1) = 4.

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Key Concepts

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

Lagrange multiplier
Lagrange multipliers are a powerful tool to handle optimization problems, especially when there is a constraint involved. Imagine you want to find the highest or lowest point of a surface, but you have to stay on a certain path. Lagrange multipliers help you navigate such situations.

Here's how it works:
  • You start with a function you want to optimize, let's call it \( f(x, y) \).
  • Then, you have a constraint that's like a rule you have to follow, noted as \( g(x, y) = 0 \).
  • The Lagrange multiplier method combines these two into one function. We introduce a new variable, \( \lambda \), and create what's called the Lagrange function: \( L(x, y, \lambda) = f(x, y) + \lambda g(x, y) \).
The value \( \lambda \) is the Lagrange multiplier. It essentially weighs how much the constraint affects the optimization of the function.

By taking this approach, we can solve for \( x, y, \) and \( \lambda \) by finding where the partial derivatives are zero. This method elegantly ties the optimization with the constraint together.
Partial derivatives
Partial derivatives are crucial in dealing with functions of more than one variable, especially in the context of optimization. Think of a partial derivative as the rate of change of a function with respect to one of its variables, while all other variables remain constant.

Here's a simple breakdown:
  • For a function \( f(x, y) \), the partial derivative with respect to \( x \) is noted as \( \frac{\partial f}{\partial x} \), showing how \( f \) changes as \( x \) changes, when \( y \) is fixed.
  • Similarly, the partial derivative with respect to \( y \) is \( \frac{\partial f}{\partial y} \).
When optimizing with constraints using Lagrange multipliers, after setting up the Lagrange function, we find the partial derivatives with respect to each variable: \( x \), \( y \), and the Lagrange multiplier \( \lambda \). Setting these derivatives to zero allows us to find points where the function might have a maximum or minimum.

These derivatives provide insights into how each variable influences the function, and when combined, they help pinpoint where the best (optimal) solution can be found.
Constraint optimization
Constraint optimization is all about finding the best result for a function given certain limitations or constraints. In real-world scenarios, you often need to optimize something—like maximizing profit or minimizing cost—while sticking to some rules.

In the context of calculus, constraints are typically expressed as equations relating the variables of the main function. For instance, in the problem we solved, the constraint was \( xy = 2 \). This constraint means any solution must satisfy the relationship between \( x \) and \( y \).

To efficiently solve such problems, we usually:
  • Start by defining the function we aim to optimize.
  • Identify and clearly express the constraints.
  • Use techniques like Lagrange multipliers to integrate the function and constraint into a single formulation.
By working through this process, we ensure that any solution found not only attempts to optimize the function but also satisfies the given constraint, achieving the optimal result under the limitations provided.

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