Problem 46 - Calculus for Biology and Medicine

Question

Let $$f(x, y)=x+y \quad(x, y) \in \mathbf{R}^{2}$$ with constraint function \(x y=1\). (a) Use Lagrange multipliers to find all local extrema. (b) Are there global extrema?

Step-by-Step Solution

Answer

Local extrema at (1, 1) with max 2; (-1, -1) with min -2. No global extrema.

Step 1: Set Up the Lagrangian

Given the function \(f(x, y) = x + y\) and the constraint \(g(x, y) = xy - 1 = 0\), we define the Lagrangian as \[ \mathcal{L}(x, y, \lambda) = x + y + \lambda (xy - 1) \] where \(\lambda\) is the Lagrange multiplier.

Step 2: Find the Critical Points Using Lagrange Multipliers

Take the partial derivatives of the Lagrangian with respect to \(x\), \(y\), and \(\lambda\) and set them equal to zero. - \( \frac{\partial \mathcal{L}}{\partial x} = 1 + \lambda y = 0 \)- \( \frac{\partial \mathcal{L}}{\partial y} = 1 + \lambda x = 0 \)- \( \frac{\partial \mathcal{L}}{\partial \lambda} = xy - 1 = 0 \)Solving the first two equations gives \( \lambda = -\frac{1}{y} \) and \( \lambda = -\frac{1}{x} \). Setting \(-\frac{1}{y} = -\frac{1}{x}\) implies \(x = y\).

Step 3: Solve the System of Equations

Since \(x = y\), substitute into the constraint \(xy = 1\);\[ x^2 = 1 \]which has solutions \(x = 1\) and \(x = -1\). Thus, the critical points are \((x, y) = (1, 1)\) and \((x, y) = (-1, -1)\).

Step 4: Evaluate the Function at the Critical Points

Evaluate \(f(x, y) = x + y\) at the critical points:- At \((1, 1)\): \( f(1, 1) = 1 + 1 = 2 \)- At \((-1, -1)\): \( f(-1, -1) = -1 - 1 = -2 \)

Step 5: Determine the Nature of Extrema and Check for Global Extrema

The function evaluations suggest:- The point \((1, 1)\) gives a local maximum of 2.- The point \((-1, -1)\) gives a local minimum of -2.For global extrema, note that if \((x, y)\) are on the hyperbola \(xy = 1\), \(x + y\) can increase or decrease without bound as \(x\) moves away from 0 (while satisfying the constraint), indicating no global extremum for \(f(x, y)\).

Key Concepts

Local Extrema

When dealing with functions of multiple variables, identifying local extrema is crucial to understanding their behavior. Local extrema are points on the graph of a function where the function achieves a local maximum or minimum value. These are essentially peaks (maximum) or troughs (minimum) when looking at the function's curve within a specific region.

In the exercise provided, we explore the function \(f(x, y) = x + y\) under the constraint that \(xy = 1\). To find local extrema, we initially set up the Lagrangian \(\mathcal{L}(x, y, \lambda) = x + y + \lambda(xy - 1)\), where \(\lambda\) is the Lagrange multiplier.

By solving the system of equations derived from the partial derivatives of the Lagrangian, we find critical points where local extrema can occur. In this case, the critical points are \((1,1)\) and \((-1,-1)\).

• At \((1, 1)\), evaluating the function gives \(f(1, 1) = 2\), which is a local maximum.
• At \((-1, -1)\), we have \(f(-1, -1) = -2\), revealing a local minimum.

These points are the local extrema because they are solutions where the Lagrangian's constraints are met, and the function's derivative indicates maximum or minimum behavior within the neighborhood.

Global Extrema

Global extrema, unlike local extrema, refer to the highest and lowest values a function can achieve over its entire domain. However, global extrema can be challenging to identify when a function is subject to specific constraints, such as in this problem where \(xy = 1\).

To determine whether global extrema exist for the function \(f(x, y) = x + y\) with the constraint, we observe the constraint hyperbola. This set of points (\(xy = 1\)) can guide us in checking the entire behavior of \(f\):

• If we let \(x\) increase to a large positive value, \(y\) would approach zero, causing \(x + y\) to grow infinitely.
• Conversely, if \(x\) and \(y\) tend towards zero from opposite sides of the constraint, their sum can decrease without bound.

These observations suggest that \(f(x, y)\) does not have a global maximum or global minimum, as \(x + y\) can continuously increase or decrease without violation of the constraint. Thus, while local extrema exist, the absence of global bounds is confirmed by the function's unrestricted growth and decay along the constraint.

Constraint Function

A constraint function in the context of optimization problems defines a condition that the solutions must satisfy. This transforms an unconstrained problem into one where the variables are limited by the constraint's equation. In our example, the constraint function is \(xy = 1\), defining a hyperbola in the plane.

Understanding the role of the constraint function is key in applying techniques like Lagrange multipliers. By embedding constraints into the optimization process, we ensure that any extrema found are valid under the given conditions. Here’s a bit more about how they work:

• The constraint \(xy = 1\) restricts our possible solutions to points that lie on the hyperbola.
• The Lagrangian multiplier \(\lambda\) allows us to incorporate the constraint directly into the derivative conditions. This ensures that the calculus considers both the function and its constraint simultaneously.

By using this method, we systematically find critical points that satisfy both the function's optimization and the constraint's conditions. This approach is elegant and powerful, enabling us to solve complex optimization problems in higher dimensions while adhering strictly to the specified constraints.