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 maximum value is \( \frac{4}{9} \) at \((1, -\frac{2}{3})\).

Step 1: Define the Lagrange Function

To apply Lagrange multipliers, define the Lagrange function, \( \mathcal{L}(x, y, \lambda) = f(x, y) + \lambda (g(x, y)) \), where \( f(x, y) = x^2 y^2 \) is the original function and \( g(x, y) = 2x - 3y - 4 = 0 \) represents the constraint. Thus, \( \mathcal{L}(x, y, \lambda) = x^2 y^2 + \lambda (2x - 3y - 4) \).

Step 2: Compute Partial Derivatives

Compute the partial derivatives of the Lagrange function with respect to \( x \), \( y \), and \( \lambda \). These are \( \frac{\partial \mathcal{L}}{\partial x} = 2xy^2 + 2\lambda \), \( \frac{\partial \mathcal{L}}{\partial y} = 2x^2 y - 3\lambda \), and \( \frac{\partial \mathcal{L}}{\partial \lambda} = 2x - 3y - 4 \).

Step 3: Set Partial Derivatives to Zero

To find the critical points, set each of the partial derivatives to zero: \( 2xy^2 + 2\lambda = 0 \), \( 2x^2 y - 3\lambda = 0 \), and \( 2x - 3y - 4 = 0 \).

Step 4: Solve the System of Equations

Solve the system of equations formed: \( xy^2 + \lambda = 0 \), \( 2x^2 y = 3\lambda \), and \( 2x - 3y = 4 \). First, express \( \lambda \) in terms of \( x \) and \( y \) using the first equation: \( \lambda = -xy^2 \). Substitute this into the second equation to get \( 2x^2 y = -3xy^2 \), simplifying to \( 2x^2 = -3xy \).

Step 5: Simplify and Solve for Variables

Simplifying \( 2x^2 = -3xy \) gives \( y = -\frac{2}{3}x \) or \( x = 0 \). Substituting into the constraint \( 2x - 3y = 4 \), when \( x = 0 \), \( y = -\frac{4}{3} \). Using \( y = -\frac{2}{3}x \), place it back into the constraint, giving \( 2x - 3(-\frac{2}{3}x) = 4 \), solving \( x = 1 \), \( y = -\frac{2}{3} \).

Step 6: Evaluate f at Critical Points

Evaluate \( f(x, y) = x^2 y^2 \) at the two discovered points. At \((0, -\frac{4}{3})\), \( f(x,y) = 0 \). At \((1, -\frac{2}{3})\), \( f(x,y) = (1)^2 (-\frac{2}{3})^2 = \frac{4}{9} \). This indicates \((1, -\frac{2}{3})\) as a max point, and \((0, -\frac{4}{3})\) produces no value or indication of a minimum.

Key Concepts

Partial Derivatives

Understanding partial derivatives is crucial when working with multivariable functions. A partial derivative measures how a function changes as one of its variables changes, while keeping all other variables constant. In the context of Lagrange multipliers, partial derivatives are used to examine the behavior of a function concerning each variable, given a constraint.

Let's consider the Lagrange function defined in our exercise, \( \mathcal{L}(x, y, \lambda) = x^2 y^2 + \lambda (2x - 3y - 4) \). To solve it, we compute the partial derivatives of \( \mathcal{L} \) with respect to each variable:

• \( \frac{\partial \mathcal{L}}{\partial x} = 2xy^2 + 2\lambda \)
• \( \frac{\partial \mathcal{L}}{\partial y} = 2x^2 y - 3\lambda \)
• \( \frac{\partial \mathcal{L}}{\partial \lambda} = 2x - 3y - 4 \)

By setting these partial derivatives to zero, we find the critical points where the function could have maxima or minima under the given constraint.

System of Equations

After determining the partial derivatives and setting them to zero, solving the resulting system of equations is the next step toward finding the extrema of the function under the given constraint.

In this exercise, the system arises from the equations:

• \( 2xy^2 + 2\lambda = 0 \)
• \( 2x^2 y - 3\lambda = 0 \)
• \( 2x - 3y - 4 = 0 \)

These equations represent the relationships between the function's variables and the constraint. Solving them involves expressing one variable in terms of others and substituting back to find specific values for each variable.

For example, we express \( \lambda \) as \( \lambda = -xy^2 \) from the first equation, and substitute into the second equation, ultimately solving to find values for \( x \) and \( y \). Solving a system of equations can be simplified by examining each equation and using algebraic manipulation to isolate variables.

Constraint Optimization

Constraint optimization refers to finding an extremum (maximum or minimum) of a function while obeying a constraint, often expressed as an equation. In problems like the given exercise, Lagrange multipliers help solve these constrained optimization problems efficiently.

The method involves creating a Lagrange function, which incorporates the original function and the constraint using a new variable \( \lambda \). We then use partial derivatives of the Lagrange function to establish critical points, dictated both by the behavior of the function and the constraint.

In this case, our constraint is \( 2x - 3y = 4 \). By finding where these conditions hold simultaneously through the system of equations, we locate possible extrema. Evaluating the original function at these points reveals which is a minimum or maximum.

Ultimately, constraint optimization allows us to find the most efficient or effective solution within defined limits, which is invaluable in fields like economics, engineering, and resource management.