Problem 40 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The function has an extremum at \( (\frac{2}{3}, \frac{-1}{3}) \) with value \( \frac{1}{3} \).

Step 1: Identify the Function and Constraint

The function to optimize is given as \( f(x, y) = x^2 - y^2 \), and the constraint is \( g(x, y) = 2x + y - 1 = 0 \). We will use the method of Lagrange multipliers to find the extremum of the function subject to this constraint.

Step 2: Formulate the Lagrangian

We formulate the Lagrangian, \( \mathcal{L}(x, y, \lambda) = f(x, y) + \lambda g(x, y) \), where \( \lambda \) is the Lagrange multiplier. Thus, \( \mathcal{L}(x, y, \lambda) = x^2 - y^2 + \lambda(2x + y - 1) \).

Step 3: Compute Partial Derivatives

We compute the partial derivatives of the Lagrangian. \[ \frac{\partial \mathcal{L}}{\partial x} = 2x + 2\lambda \]\[ \frac{\partial \mathcal{L}}{\partial y} = -2y + \lambda \]\[ \frac{\partial \mathcal{L}}{\partial \lambda} = 2x + y - 1 \]

Step 4: Set Partial Derivatives to Zero

Set each of the partial derivatives to zero to find critical points. We have three equations:1. \( 2x + 2\lambda = 0 \)2. \( -2y + \lambda = 0 \)3. \( 2x + y - 1 = 0 \) (this is the constraint)

Step 5: Solve the System of Equations

From equation 1, solve for \( \lambda \):\( \lambda = -x \).From equation 2, express \( y \) in terms of \( \lambda \):\( y = \frac{\lambda}{2} \).Substitute \( \lambda = -x \) into \( y \):\( y = \frac{-x}{2} \).Using equation 3 (the constraint):\( 2x + \left(-\frac{x}{2}\right) = 1 \), which simplifies to \( \frac{3x}{2} = 1 \).Solve for \( x \):\( x = \frac{2}{3} \).Substitute \( x = \frac{2}{3} \) back into the expression for \( y \):\( y = \frac{-1}{3} \).

Step 6: Evaluate the Function at Critical Points

We evaluate the function at the critical point \((x, y) = \left(\frac{2}{3}, \frac{-1}{3}\right)\) to find the extremum.\[ f\left(\frac{2}{3}, \frac{-1}{3}\right) = \left(\frac{2}{3}\right)^2 - \left(-\frac{1}{3}\right)^2 = \frac{4}{9} - \frac{1}{9} = \frac{3}{9} = \frac{1}{3} \].This is the value of the function at the critical point.

Key Concepts

Constrained Optimization

Constrained optimization is a technique used to find the maximum or minimum of a function when there are restrictions or conditions that the solutions must satisfy. Imagine you're shopping with a budget; your spending limit is the constraint. In mathematics, constraints are given as equations or inequalities. Here, the goal is to optimize a function, known as the objective function, while respecting these limitations.
In our example, the function is \(f(x, y) = x^2 - y^2\) and the constraint is \(2x + y = 1\). The constraint drastically changes the solution because it restricts the region where we search for the optimum. Rather than looking everywhere, we only focus on points that fit the constraint.

Partial Derivatives

Partial derivatives are crucial in the process of optimization. They represent how a function changes as each of the variables changes, while the others remain constant. Imagine navigating a mountain; partial derivatives tell you how steep the slope is if you only move east or north, separately.
For the Lagrangian \(\mathcal{L}(x, y, \lambda) = x^2 - y^2 + \lambda(2x + y - 1)\), we calculate its partial derivatives with respect to each variable:

• For \(x\): \(\frac{\partial \mathcal{L}}{\partial x} = 2x + 2\lambda\)
• For \(y\): \(\frac{\partial \mathcal{L}}{\partial y} = -2y + \lambda\)
• For \(\lambda\): \(\frac{\partial \mathcal{L}}{\partial \lambda} = 2x + y - 1\)

Each derivative gives insight into how to adjust each variable to improve the value of the objective function while adhering to the constraint.

Critical Points

The concept of critical points in optimization is key because these are the points where a function could have a max or min. Think of these as the peaks or valleys as you hike a mountain. To find these points mathematically, we solve systems of equations derived by setting partial derivatives equal to zero.
In our case, the equations are:

• \(2x + 2\lambda = 0\)
• \(-2y + \lambda = 0\)
• \(2x + y - 1 = 0\)

These equations stem from the condition that the slope in any direction on the constraint is zero, reflecting that you are neither climbing up nor slipping down. Solving gives the critical point \((x, y) = (\frac{2}{3}, \frac{-1}{3})\).

Extremum of Functions

An extremum of a function indicates its highest or lowest output under given conditions. This is like finding the tallest building or the deepest valley in a city, but with respect to function values and constraints. Once we identify a function’s critical points, we substitute these values back into the original function to determine whether these points are maxima or minima.
For the function \(f(x, y) = x^2 - y^2\), after finding the critical point \((\frac{2}{3}, \frac{-1}{3})\), we substitute it in: \[ f\left(\frac{2}{3}, \frac{-1}{3}\right) = \left(\frac{2}{3}\right)^2 - \left(-\frac{1}{3}\right)^2 = \frac{4}{9} - \frac{1}{9} = \frac{3}{9} = \frac{1}{3} \]
This value is the function’s extremum under the constraint, representing the function’s optimal output in that constrained region.