Problem 39 - Calculus for Biology and Medicine

Question

Use Lagrange multipliers to find the maxima and minima of the functions under the given constraints. $$ f(x, y)=x y ; x^{2}+4 y^{2}=1 $$

Step-by-Step Solution

Answer

Maxima: \( (\frac{1}{\sqrt{2}}, \frac{1}{2\sqrt{2}}), (-\frac{1}{\sqrt{2}}, -\frac{1}{2\sqrt{2}}) \); Minima: \( (\frac{1}{\sqrt{2}}, -\frac{1}{2\sqrt{2}}), (-\frac{1}{\sqrt{2}}, \frac{1}{2\sqrt{2}}) \).

Step 1: Function and Constraint Definition

First, identify the function to optimize and the constraint. We need to optimize the function \( f(x, y) = x y \) subject to the constraint \( g(x, y) = x^2 + 4y^2 - 1 = 0 \).

Step 2: Set Up Lagrange Function

Introduce the Lagrange multiplier \( \lambda \) and set up the Lagrange function: \( \mathcal{L}(x, y, \lambda) = xy + \lambda (1 - x^2 - 4y^2) \).

Step 3: Compute Partial Derivatives

Find the partial derivatives of \( \mathcal{L} \) with respect to \( x \), \( y \), and \( \lambda \).- \( \frac{\partial \mathcal{L}}{\partial x} = y - 2\lambda x = 0 \)- \( \frac{\partial \mathcal{L}}{\partial y} = x - 8\lambda y = 0 \)- \( \frac{\partial \mathcal{L}}{\partial \lambda} = 1 - x^2 - 4y^2 = 0 \)

Step 4: Solve Partial Derivative Equations

Use the equations from step 3 to form a system of equations:1. \( y = 2\lambda x \)2. \( x = 8\lambda y \)3. \( x^2 + 4y^2 = 1 \).From the first two equations: \( y = 2\lambda x \) and \( x = 8\lambda y \), substitute one into the other to express \( \lambda \).

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

Substituting \( y = 2\lambda x \) into \( x = 8\lambda y \) gives: \( x = 8\lambda(2\lambda x) \) leading to \( 1 = 16\lambda^2 \) or \( \lambda = \pm \frac{1}{4} \). For \( \lambda = \frac{1}{4} \), solve \( y = \frac{1}{2}x \). Substitute into constraint to find points. Do similarly for \( \lambda = -\frac{1}{4} \).

Step 6: Find Critical Points

For \( \lambda = \frac{1}{4} \): Solve \( x^2 + 4\left(\frac{1}{2}x\right)^2 = 1 \) resulting in \( x^2 + x^2 = 1 \) which simplifies to \( x = \pm \frac{1}{\sqrt{2}} \), thus \( y = \pm \frac{1}{2\sqrt{2}} \). Points: \( \left( \frac{1}{\sqrt{2}}, \frac{1}{2\sqrt{2}} \right) \) and \( \left( -\frac{1}{\sqrt{2}}, -\frac{1}{2\sqrt{2}} \right) \). Similarly for \( \lambda = -\frac{1}{4} \), find points \( \left( \frac{1}{\sqrt{2}}, -\frac{1}{2\sqrt{2}} \right) \) and \( \left( -\frac{1}{\sqrt{2}}, \frac{1}{2\sqrt{2}} \right) \).

Step 7: Evaluate Function Values at Critical Points

Evaluate \( f(x, y) = xy \) at the critical points:1. \( f\left( \frac{1}{\sqrt{2}}, \frac{1}{2\sqrt{2}} \right) = \frac{1}{4} \)2. \( f\left( -\frac{1}{\sqrt{2}}, -\frac{1}{2\sqrt{2}} \right) = \frac{1}{4} \)3. \( f\left( \frac{1}{\sqrt{2}}, -\frac{1}{2\sqrt{2}} \right) = -\frac{1}{4} \)4. \( f\left( -\frac{1}{\sqrt{2}}, \frac{1}{2\sqrt{2}} \right) = -\frac{1}{4} \).

Step 8: Determine Maxima and Minima

Identify maxima and minima:- Maximum values at points \( \left( \frac{1}{\sqrt{2}}, \frac{1}{2\sqrt{2}} \right) \) and \( \left( -\frac{1}{\sqrt{2}}, -\frac{1}{2\sqrt{2}} \right) \) with \( f(x, y) = \frac{1}{4} \).- Minimum values at points \( \left( \frac{1}{\sqrt{2}}, -\frac{1}{2\sqrt{2}} \right) \) and \( \left( -\frac{1}{\sqrt{2}}, \frac{1}{2\sqrt{2}} \right) \) with \( f(x, y) = -\frac{1}{4} \).

Key Concepts

Maxima and Minima

In calculus, finding the maxima and minima of a function involves identifying the largest and smallest values the function can take within a given range or under specific constraints. These points are essential because they can represent optimal solutions or critical points in real-world problems such as physics, engineering, and economics.

**Maxima** refers to the highest point(s) on the graph of the function, while **minima** are the lowest point(s). They can be of two types:

• Local maxima/minima: Points where the function yields a maximum or minimum value within a small interval around those points.
• Global maxima/minima: Points where the function obtains a maximum or minimum value over its entire domain.

In the context of constraint optimization, maxima and minima are found under given constraints. For instance, in the original exercise, the task was to find these values for the function \( f(x, y) = xy \) under the constraint \( x^2 + 4y^2 = 1 \). This means we're looking for the largest and smallest product of \( x \) and \( y \) while remaining on the ellipse defined by the constraint. Using Lagrange multipliers, we narrowed down to critical points, then evaluated the function at these points to find two maxima at \( \frac{1}{4} \) and two minima at \(-\frac{1}{4} \).

Constraint Optimization

Constraint optimization is a method used when a problem imposes restrictions or rules in addition to aiming at optimizing a function's output. Often, these constraints can be expressed as an equation or inequality involving the variables of the function.

Consider the original example: the function \( f(x, y) = xy \) was optimized with the condition \( x^2 + 4y^2 = 1 \). This type of problem is common in real-world situations where resources are limited or conditions apply. For instance, a company trying to maximize profit (the objective function) may also need to comply with environmental regulations (the constraint).

• The key to handling constraint optimization problems effectively is to capture these limitations mathematically and solve them using appropriate techniques.
• Lagrange multipliers are an essential tool in these scenarios, providing a systematic way to include the constraints in our calculation process.

The basic idea is to transform the problem into solving the Lagrange function \( \mathcal{L}(x, y, \lambda) \), which incorporates the original function and the constraint. This new function treats the constraint as another part of the system to be solved, allowing us to find the optimal solution that respects the constraint.

Partial Derivatives

Partial derivatives are fundamental in multivariable calculus, allowing us to examine how a function changes as one of the multiple variables changes, while others are kept constant. They are particularly useful in analyzing the behavior of functions with more than one variable and are crucial in the method of Lagrange multipliers.

When we construct the Lagrange function for a constraint optimization problem, computing the partial derivatives is a critical step. It involves taking derivatives with respect to each variable, including the Lagrange multiplier, to set up a system of equations
To solve these equations, you extract critical points, which can potentially be maxima or minima of the function.
Here is how the partial derivatives work in our exercise:

• For the Lagrange function \( \mathcal{L}(x, y, \lambda) = xy + \lambda (1 - x^2 - 4y^2) \), we calculated the partial derivatives with respect to \( x \), \( y \), and \( \lambda \):
• Setting these derivatives equal to zero provides the equations necessary to solve the optimization problem.

Partial derivatives give us the essential information about the function's slopes and make constraint optimization feasible, helping identify where and how the function's rate of change hits zero, marking potential points of interest.