Problem 8 - Applied Calculus

Question

Use Lagrange multipliers to find the maximum or minimum values of \(f(x, y)\) subject to the constraint. $$ f(x, y)=x^{2}+y, \quad x^{2}-y^{2}=1 $$

Step-by-Step Solution

Answer

Both the maximum and minimum value is \(\frac{1}{2}\), at points \((\frac{\sqrt{5}}{2}, -\frac{1}{2})\) and \((-\frac{\sqrt{5}}{2}, -\frac{1}{2})\).

Step 1: Understand Lagrange multipliers

Lagrange multipliers is a strategy for finding the maxima and minima of a function subject to equality constraints. Given a function \(f(x, y)\) and a constraint \(g(x, y) = 0\), we solve the problem by finding methods to compute gradients, where the gradients are parallel: \( abla f = \lambda abla g \).

Step 2: Define the Gradients

Write down the partial derivatives for both the function and the constraint. For \( f(x, y) = x^2 + y \), we have \( \frac{\partial f}{\partial x} = 2x \) and \( \frac{\partial f}{\partial y} = 1 \). For the constraint \( g(x, y) = x^2 - y^2 - 1 \), we have \( \frac{\partial g}{\partial x} = 2x \) and \( \frac{\partial g}{\partial y} = -2y \).

Step 3: Set Up Equations with Lagrange Multiplier

Set up the equations using the Lagrange multiplier \( \lambda \). This gives us two equations: \( 2x = \lambda 2x \) and \( 1 = \lambda (-2y) \). Additionally, we have the constraint equation \( x^2 - y^2 = 1 \).

Step 4: Solve the Equations

From \( 2x = \lambda 2x \), if \(x eq 0\), then \( \lambda = 1\). From the equation \( 1 = \lambda (-2y) \), substituting \( \lambda = 1\) gives \( 1 = -2y \Rightarrow y = -\frac{1}{2} \). With these, use the constraint \( x^2 - y^2 = 1 \) to find \( x^2 - ( -\frac{1}{2})^2 = 1 \Rightarrow x^2 - \frac{1}{4} = 1 \Rightarrow x^2 = \frac{5}{4} \Rightarrow x = \pm \frac{\sqrt{5}}{2} \).

Step 5: Evaluate the Objective Function

Substitute \( x = \frac{\sqrt{5}}{2} \) or \( x = -\frac{\sqrt{5}}{2} \) and \( y = -\frac{1}{2} \) back into \( f(x, y) = x^2 + y \) to find the value of the objective function. Calculate: \( f(\frac{\sqrt{5}}{2}, -\frac{1}{2}) = (\frac{\sqrt{5}}{2})^2 - \frac{1}{2}\) and \( f(-\frac{\sqrt{5}}{2}, -\frac{1}{2}) = (\frac{\sqrt{5}}{2})^2 - \frac{1}{2}\). Both calculations yield \( \frac{1}{2} \).

Step 6: Conclusion

The maximum and minimum values of the function \( f(x, y) = x^2 + y \) subject to the constraint \( x^2 - y^2 = 1 \) are both equal to \( \frac{1}{2} \), occurring at the points \( (\frac{\sqrt{5}}{2}, -\frac{1}{2}) \) and \( (-\frac{\sqrt{5}}{2}, -\frac{1}{2}) \).

Key Concepts

Objective Function

The objective function is essentially the mathematical expression that we want to optimize, either by maximizing or minimizing its value. It serves as a representation of the problem we're trying to solve. In the context of Lagrange multipliers, the objective function, denoted as \( f(x, y) \), is the target of our optimization.In the problem above, our objective function is \( f(x, y) = x^2 + y \). Here, each variable \( x \) and \( y \) has a specific role:

• \( x^2 \): Represents a quadratic term, meaning it affects the overall function value more significantly as \( x \) increases or decreases.
• \( y \): Adds a linear component, complementing the quadratic portion.

The objective function plays a crucial role because

• It defines the ultimate goal of our problem-solving process.
• Its form (here, a mix of quadratic and linear) affects how we interpret results and constraints.

Analyzing the objective function helps us determine what we're optimizing and ensures we choose the appropriate methods to solve the problem efficiently.

Equality Constraint

An equality constraint is a condition that the solution to an optimization problem must satisfy. It restricts the set of feasible solutions to a smaller region within which we must find the extrema (maximum or minimum values) of the objective function.For our problem, the constraint is \( x^2 - y^2 = 1 \). Here's what this means:

• It is an equation that must hold true for all valid pairs of \( x \) and \( y \) when optimizing the objective.
• The constraint often defines a curve or surface, depending on the dimensionality of the variables. Here, it shapes the feasible region in the \( xy \)-plane.

The equality constraint in Lagrange multipliers leads us to consider the restricted optimization method, allowing us to explore viable solutions while maintaining adherence to the defined limits. This balances how we search for optimal points, integrating both our objective and boundaries smoothly.

Partial Derivatives

Partial derivatives are essential in examining how a function changes as we adjust one variable while keeping the other constant. They are pivotal in the use of Lagrange multipliers as they give us insight into the slope or rate of change in each direction of our function's input variables.In the provided exercise, partial derivatives are computed for both the objective function \( f(x, y) = x^2 + y \) and the constraint \( g(x, y) = x^2 - y^2 - 1 \):

• \( \frac{\partial f}{\partial x} = 2x \): Indicates the rate of change of \( f \) with respect to \( x \).
• \( \frac{\partial f}{\partial y} = 1 \): The change in \( f \) per unit change in \( y \).
• \( \frac{\partial g}{\partial x} = 2x \) and \( \frac{\partial g}{\partial y} = -2y \): These help in constructing the constraint's gradient.

The gradients of these derivatives lead us to create and solve the Lagrange equations:

• By equating them to each other (adjusted by the multiplier \( \lambda \)), we establish the system that aligns the function's gradients, essential for finding extrema while respecting the constraint.

Thus, partial derivatives not only articulate change and slope but pave the way for balanced optimization grounded by Lagrange's ingenious method.