Problem 47 - Calculus Early Transcendentals

Question

Absolute maximum and minimum values Find the absolute maximum and minimum values of the following functions over the given regions \(R\). Use Lagrange multipliers to check for extreme points on the boundary. $$f(x, y)=x^{2}-4 y^{2}+x y ; R=\left\\{(x, y): 4 x^{2}+9 y^{2} \leq 36\right\\}$$

Step-by-Step Solution

Answer

$$ Answer: The absolute maximum value is $\frac{1008}{25}$ and the absolute minimum value is $-\frac{1648}{25}$.

Step 1: Find the gradient of the function

Find the partial derivatives of f with respect to x and y: $$\frac{\partial f}{\partial x} = 2x + y$$ $$\frac{\partial f}{\partial y} = -8y + x$$

Step 2: Find the critical points of the function

A critical point occurs when the gradient of f is equal to 0. Set up the following system of equations and solve for x and y: $$2x + y = 0$$ $$-8y + x = 0$$ The solution to this system is: $$x = 16/5$$ $$y = 32/5$$

Step 3: Set up the Lagrange multiplier problem

Define a function g representing the constraint in the region R: $$g(x, y) = 4x^2 + 9y^2 - 36 = 0$$ The Lagrange multiplier equation is: $$\nabla f(x, y) = \lambda \nabla g(x, y)$$ So, we have the following system of equations for x, y, and λ: $$2x + y = 4\lambda x$$ $$-8y + x = 18\lambda y$$ $$4x^2 + 9y^2 = 36$$

Step 4: Solve the Lagrange multiplier equations

Solve the system of equations for x, y, and λ. There will be four solutions: $$(x_1, y_1) = \left(\frac{18}{5}, -\frac{24}{5}\right), \quad \lambda_1 = -\frac{1}{10}$$ $$(x_2, y_2) = \left(-\frac{18}{5}, -\frac{24}{5}\right), \quad \lambda_2 = \frac{1}{10}$$ $$(x_3, y_3) = \left(\frac{18}{5}, \frac{24}{5}\right), \quad \lambda_3 = -\frac{11}{45}$$ $$(x_4, y_4) = \left(-\frac{18}{5}, \frac{24}{5}\right), \quad \lambda_4 = \frac{11}{45}$$

Step 5: Evaluate the function at each solution and the critical point

Evaluate f at each of the four solutions found in Step 4 and the critical point found in Step 2: $$f\left(\frac{16}{5}, \frac{32}{5}\right) = -\frac{1648}{25}$$ $$f\left(\frac{18}{5}, -\frac{24}{5}\right) = \frac{1008}{25}$$ $$f\left(-\frac{18}{5}, -\frac{24}{5}\right) = \frac{1008}{25}$$ $$f\left(\frac{18}{5}, \frac{24}{5}\right) = -\frac{1008}{25}$$ $$f\left(-\frac{18}{5}, \frac{24}{5}\right) = -\frac{1008}{25}$$

Step 6: Identify the maximum and minimum values

From the evaluations in Step 5, we can see that: $$\text{Maximum value:} \quad \frac{1008}{25}$$ $$\text{Minimum value:} \quad -\frac{1648}{25}$$

Key Concepts

Gradient

To find the extreme points of a function using Lagrange multipliers, the first step is to determine the gradient. The gradient, represented by \( abla f \), is a vector composed of the partial derivatives of the function. For any function \( f(x, y) \), the gradient offers a direction along which the function increases most rapidly.

• For the function \( f(x, y) = x^2 - 4y^2 + xy \), the gradient is \( \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \).
• Partial derivative \( \frac{\partial f}{\partial x} = 2x + y \) indicates how \( f \) changes with respect to \( x \).
• Partial derivative \( \frac{\partial f}{\partial y} = -8y + x \) indicates how \( f \) changes with respect to \( y \).

Understanding the gradient helps us investigate where the function doesn't change, as those points can be potential critical points.

Critical Points

Critical points occur where the gradient of a function equals zero. These are points where the function might achieve local maxima or minima. To find these, solve for \( x \) and \( y \) such that each component of the gradient is zero.
For our example:

• Set \( 2x + y = 0 \) and \( -8y + x = 0 \).
• Solving this system of equations gives \( x = \frac{16}{5} \) and \( y = \frac{32}{5} \).

These coordinates are the critical point, where the potential extremity happens within the interior of the constraint region. Critical points must always be evaluated in conjunction with boundary checks to ensure absolute extremity.

Constraint Equations

In optimization problems involving boundaries, constraint equations define the scope within which a function needs to be optimized. The constraint determines the feasible region.
For instance, in our problem:

• The constraint in region \( R \) is given by \( g(x, y) = 4x^2 + 9y^2 - 36 = 0 \).
• This depicts an ellipse, and our task is to explore his boundary as well as areas within.

In Lagrange multipliers, we incorporate the constraint into our problem by introducing a multiplier \( \lambda \), leading to the equation \( abla f = \lambda abla g \). This setup indicates where the directional changes in \( f \) mirror those of the boundary, hinting at potential extrema.

Extreme Points

Extreme points are where a function attains its highest or lowest value under given conditions. To find these, evaluate not only the critical points but also points on the boundary of the constraint region. Using the Lagrange multiplier method provides such potential boundary points.
Consider:

• The Lagrange system from our problem gave four boundary solutions: \( \left(\frac{18}{5}, -\frac{24}{5}\right) \), \( \left(-\frac{18}{5}, -\frac{24}{5}\right) \), \( \left(\frac{18}{5}, \frac{24}{5}\right) \), and \( \left(-\frac{18}{5}, \frac{24}{5}\right) \).
• Evaluating the original function \( f \) at these points gives potential maximum and minimum values.

Ultimately, by comparing the function values at all discovered points, including critical and boundary points, we ascertain which holds the absolute maximum and minimum, leading to a comprehensive solution to the problem.