Problem 47 - Calculus for Biology and Medicine

Question

Let $$f(x, y)=x+y$$ with constraint function $$\frac{1}{x}+\frac{1}{y}=1, x \neq 0, y \neq 0$$ (a) Use Lagrange multipliers to find all local extrema. (b) Are there global extrema?

Step-by-Step Solution

Answer

Local extrema exist at \( (2, 2) \) with \( f(2, 2)=4 \). No global extrema due to constraints.

Step 1: Understand the Problem

We need to find the extrema of the function \( f(x, y) = x + y \) subject to the constraint \( \frac{1}{x} + \frac{1}{y} = 1 \) using the method of Lagrange multipliers.

Step 2: Set Up Lagrangian

Define the Lagrangian function as \( \mathcal{L}(x, y, \lambda) = x + y + \lambda \left( \frac{1}{x} + \frac{1}{y} - 1 \right) \). Here, \( \lambda \) is the Lagrange multiplier.

Step 3: Find Partial Derivatives

Calculate the partial derivatives:\- \( \frac{\partial \mathcal{L}}{\partial x} = 1 - \lambda \frac{1}{x^2} \)\- \( \frac{\partial \mathcal{L}}{\partial y} = 1 - \lambda \frac{1}{y^2} \)\- \( \frac{\partial \mathcal{L}}{\partial \lambda} = \frac{1}{x} + \frac{1}{y} - 1 \)

Step 4: Solve Partial Derivative Equations

Set each of the derivatives equal to zero:1. \( 1 - \lambda \frac{1}{x^2} = 0 \)2. \( 1 - \lambda \frac{1}{y^2} = 0 \)3. \( \frac{1}{x} + \frac{1}{y} = 1 \)

Step 5: Solve for \( \lambda \) and Variables

From equations (1) and (2), solve for \( \lambda \):\- \( \lambda = x^2 \)\- \( \lambda = y^2 \)\So, \( x^2 = y^2 \), or \( x = y \) or \( x = -y \).

Step 6: Evaluate Constraints

Substitute \( x = y \) into the constraint \( \frac{1}{x} + \frac{1}{y} = 1 \) gives \( \frac{2}{x} = 1 \), or \( x = 2 \), \( y = 2 \). For \( x = -y \), the equation simplifies to no real solution for positive integers.

Step 7: Evaluate Function at Critical Points

Evaluate \( f(x, y) \): \- For \( x = 2, y = 2 \): \( f(2, 2) = 4 \). \No valid points arise for \( x = -y \) under the non-zero, positive constraint.

Step 8: Determine Global Extrema

The constrained equation \( \frac{1}{x} + \frac{1}{y} = 1 \) shows that solutions are positive. Therefore, \( f(x, y) \) does not have a bounded maximum or minimum in both directions, hence no global extrema.

Key Concepts

Extrema

Extrema are basically the highest or lowest points on a graph. They include both minima (lowest) and maxima (highest) values. When dealing with functions in two variables, such as in our problem, you're looking at surfaces in three-dimensional space.
These points where surfaces peak or dip are what we call extrema. In the context of the problem, finding extrema involves looking at how the function behaves when we apply certain constraints.

• Local extrema: These are the peaks and dips in specific regions or domains of the function.
• Global extrema: These refer to the absolute highest or lowest points for the entire domain that the function can have.

The primary goal here is to determine these points using the method of Lagrange multipliers. For this particular exercise, the function indeed has a local extremum when both x and y equal 2. However, the constraints prevent the function from having any global extrema because they restrict x and y to certain positive values only.

Constraint Optimization

Constraint optimization is a powerful concept in calculus, particularly when you need to find the extremum of a function that has to satisfy certain conditions or constraints. Here, the function you want to optimize is usually subject to some restrictions.
In simpler words, while you seek the highest or lowest points (extrema), you need to keep the function within the confines of the constraints. In our case, the constraint is given by the equation \( \frac{1}{x} + \frac{1}{y} = 1 \). The aim is to optimize the function \( f(x, y) = x + y \) under this specific limit. Instead of exploring the entire space where the function exists, constraint optimization focuses only on the feasible region defined by \( x \) and \( y \) that satisfy the constraint.

• Identifying feasible points: These are the values of \( x \) and \( y \) that comply with the constraint.
• Finding optima within these: This is where mathematical tools help determine whether these points represent true maxima or minima.

The tool we use for this in calculus is often Lagrange multipliers, which simplifies the process significantly by turning the constrained problem into a sequence of equations.

Partial Derivatives

Partial derivatives are an extension of the concept of derivatives to higher dimensions. When working with functions of multiple variables, understanding how small changes in any one variable affect the function is crucial. This is where partial derivatives come into play.
They tell you the rate at which the function changes as just one of the variables is varied. For the function \( f(x, y) = x + y \) from the problem, we use partial derivatives to isolate how changes in \( x \) or \( y \) alone affect \( f(x, y) \).

• Computing partial derivatives: You differentiate the function with respect to one variable while keeping the other constant.
• Applying in optimization: These derivatives inform us about the slope of the function in the direction of each variable.

In the stepwise solution, you'll notice that partial derivatives are crucial for setting up the equations that you need to solve when using Lagrange multipliers. They essentially form the basis for making good judgments about how the function behaves locally, allowing optimization within constraints to be efficient.