Problem 44 - Calculus for Biology and Medicine

Question

Use Lagrange multipliers to find the maxima and minima of the functions under the given constraints. \(f(x, y)=x^{2} y^{2} ; 2 x-3 y=4\)

Step-by-Step Solution

Answer

The critical point (1, -2/3) gives the minimum value of the function, \( \frac{4}{9} \), under the constraint.

Step 1: Set Up Lagrange Function

First, identify the function to maximize or minimize, \( f(x, y) = x^2 y^2 \), and the constraint, \( g(x, y) = 2x - 3y - 4 = 0 \). Define the Lagrange function, \( \mathcal{L}(x, y, \lambda) = x^2 y^2 + \lambda (2x - 3y - 4) \).

Step 2: Find Partial Derivatives

Next, calculate the partial derivatives of the Lagrange function: \( \frac{\partial \mathcal{L}}{\partial x} = 2xy^2 + 2\lambda \), \( \frac{\partial \mathcal{L}}{\partial y} = 2x^2y - 3\lambda \), and \( \frac{\partial \mathcal{L}}{\partial \lambda} = 2x - 3y - 4 \).

Step 3: Set Partial Derivatives to Zero

Set each of the partial derivatives equal to zero to find critical points: \[ 2xy^2 + 2\lambda = 0, \quad 2x^2y - 3\lambda = 0, \quad 2x - 3y = 4 \].

Step 4: Solve for \(\lambda\) and \(x,y\)

From \( 2xy^2 + 2\lambda = 0 \), solve for \( \lambda \): \( \lambda = -xy^2 \). From \( 2x^2y - 3\lambda = 0 \), plug in \( \lambda = -xy^2 \) to get: \( 2x^2y + 3xy^2 = 0 \) leading to \( xy(2x + 3y) = 0 \). Because \( xy = 0 \) doesn't satisfy the constraint, solve \( 2x + 3y = 0 \).

Step 5: Solve System of Equations

Using \( 2x + 3y = 0 \) with \( 2x - 3y = 4 \), solve the linear equations. Substitute \( x = -\frac{3}{2}y \) from the first equation into the second equation to get: \[ 2\left(-\frac{3}{2}y\right) - 3y = 4 \] \( \Rightarrow -3y - 3y = 4 \) \( \Rightarrow -6y = 4 \) \( \Rightarrow y = -\frac{2}{3} \). Use \( x = -\frac{3}{2}y \) to find \( x = 1 \).

Step 6: Calculate Function Values

Plug \( x = 1 \), \( y = -\frac{2}{3} \) into the original function \( f(x, y) = x^2y^2 \) to get \[ f(1, -\frac{2}{3}) = 1^2 \left(-\frac{2}{3}\right)^2 = \frac{4}{9}. \] There is only one critical point, so this must be either a maximum or a minimum.

Key Concepts

Constraint Optimization

Constraint optimization is a method used to find the highest or lowest values of a function within given constraints. These constraints are conditions or limits set on the problem, which can be equality or inequality constraints. Think of it as trying to maximize or minimize a function but only within a certain region limited by these constraints.
In our exercise, we dealt with the function given as \(f(x, y) = x^2 y^2\). The constraint here is defined by the equation \(2x - 3y = 4\). This linear equation describes a line in the coordinate plane, and we aim to find points along this line that either maximize or minimize our function \(f(x, y)\).
The process involves setting up a Lagrange function that incorporates both the function and the constraint. The power of constraint optimization lies in its ability to efficiently find solutions that respect given conditions. It is a critical concept in optimization used across various fields in science and engineering.

Critical Points

Critical points are special values of the variables in our problem that can potentially represent maximum, minimum, or saddle points of the function. They are found by solving for where the first derivatives of the Lagrange function are zero.
In our exercise, we started by taking partial derivatives of the Lagrange function \(\mathcal{L}(x, y, \lambda)\), which gives equations that allow us to locate these critical points. These derivatives are set to zero to form a system of equations. Solving these equations reveals the values of \(x\), \(y\), and \(\lambda\) that define our critical points.
Once we identify these points, inserting them back into the original function lets us determine if they represent a maximum, minimum, or neither. In our situation, after solving, the critical point was \(x = 1\) and \(y = -\frac{2}{3}\), and checking the value of the function at this point, we determined the minimum or maximum nature of this point.

Partial Derivatives

Partial derivatives indicate how a function changes as one of the variables is altered while keeping the others constant. They are a fundamental part of calculus that helps in understanding how functions behave.
In our problem, we found the partial derivatives of the Lagrange function \(\mathcal{L}(x, y, \lambda)\) with respect to \(x\), \(y\), and \(\lambda\). These derivatives are essential for the application of the Lagrange multiplier method, as they are used to identify the critical points of the function by solving where each derivative equals zero:

• The derivative with respect to \(x\), \(\frac{\partial \mathcal{L}}{\partial x} = 2xy^2 + 2\lambda\), helps track changes as \(x\) is varied.
• The derivative with respect to \(y\), \(\frac{\partial \mathcal{L}}{\partial y} = 2x^2y - 3\lambda\), shows changes as \(y\) is altered.
• The derivative with respect to \(\lambda\), \(\frac{\partial \mathcal{L}}{\partial \lambda} = 2x - 3y - 4\), fulfills the constraint condition.

Understanding how to compute and interpret these partial derivatives is critical in solving optimization problems using methods like Lagrange multipliers.