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. $$f(x, y)=2 x^{2}+y^{2}+2 x-3 y ; R=\left\\{(x, y): x^{2}+y^{2} \leq 1\right\\}$$

Short answer

Answer: The absolute maximum value of the function is 6, and the absolute minimum value is \(-\frac{13}{3}\).

Step-by-step solution

Find critical points inside the region R

To find the critical points of the function \(f(x, y)\) inside the region \(R\), we need to calculate the partial derivatives of the function with respect to \(x\) and \(y\) and set them equal to zero: $$\frac{\partial f}{\partial x} = 4x + 2$$ $$\frac{\partial f}{\partial y} = 2y - 3$$ Now, set these partial derivatives equal to zero and solve for \(x\) and \(y\): $$4x + 2 = 0 \Rightarrow x = -\frac{1}{2}$$ $$2y - 3 = 0 \Rightarrow y = \frac{3}{2}$$ However, since the point \((-\frac{1}{2}, \frac{3}{2})\) lies outside the region \(R\), there are no critical points inside the region.

Setup the Lagrange multiplier condition

The boundary of the region \(R\) is the circle \(x^2 + y^2 = 1\). We will use Lagrange multipliers to find the critical points on the boundary. The Lagrange multiplier condition is given by the following equations: $$\nabla f(x,y) = \lambda \nabla g(x,y)$$ Where \(g(x,y) = x^2 + y^2 - 1\) is the constraint equation. This gives us the following system of equations: $$4x + 2 = 2\lambda x$$ $$2y - 3 = 2\lambda y$$ $$x^2 + y^2 = 1$$

Solve the system of equations

We need to find the values of \(x\), \(y\), and \(\lambda\) that satisfy the above system of equations. We can divide the first equation by 2 and rearrange terms to get: $$\lambda = 2 - \frac{1}{x}$$ Similarly, we can divide the second equation by 2 and rearrange terms: $$\lambda = \frac{3}{2} - \frac{1}{y}$$ Now, we can equate these expressions for \(\lambda\): $$2 - \frac{1}{x} = \frac{3}{2} - \frac{1}{y}$$ We can now solve for \(y\) in terms of \(x\) (or vice versa): $$y = \frac{x(3x - 2)}{x-2}$$ Now, substitute this expression for \(y\) into the constraint equation \(x^2 + y^2 = 1\): $$x^2 + (\frac{x(3x - 2)}{x-2})^2 = 1$$ Solving this quadratic equation, we get two solutions for \(x\): $$x_a = \frac{1}{3}, x_b = -1$$ Corresponding \(y\) values can be calculated using the expression for \(y\) in terms of \(x\): $$y_a = \frac{8}{3}, y_b = 1$$ So, we have two critical points on the boundary of the region \(R\): \((\frac{1}{3}, \frac{8}{3})\) and \((-1, 1)\).

Determine the extreme values

Plug the critical points found in Step 3 back into the function \(f(x,y)\) to determine their corresponding function values: $$f(\frac{1}{3}, \frac{8}{3}) = -\frac{13}{3}$$ $$f(-1, 1) = 6$$ Now, we can compare these function values to find the absolute maximum and minimum values of the function over the region \(R\): The absolute maximum value is \(f(-1, 1) = 6\). The absolute minimum value is \(f(\frac{1}{3}, \frac{8}{3}) = -\frac{13}{3}\).

Key concepts

Absolute Maximum and Minimum

When you're dealing with optimization problems, finding the absolute maximum and minimum values of a function is often crucial. These values help identify the highest and lowest points that a function can attain over a specific region, known as the boundary. In the given exercise, the region is defined as a circle with the equation \(x^2 + y^2 \leq 1\). To find these extreme values, you'll want to analyze both the interior of the region, where partial derivatives are set to zero, and the boundary, typically requiring Lagrange multipliers. The absolute maximum is the highest value reached within and on the edge of the region, while the absolute minimum is the lowest.Key steps involve:

• Calculating the partial derivatives to find critical points inside the region.
• Utilizing Lagrange multipliers to check for extreme points on the boundary.
• Comparing the resultant values to determine the maximum and minimum.

In this particular example, the absolute maximum and minimum values found were 6 and -\(\frac{13}{3}\) respectively.

Extreme Points

Extreme points in a function are where a function takes its highest or lowest value at that point, over a defined region. These points can occur at the interior of the region or on its boundary. Finding these points requires solving for when the gradients, or slopes, of the function equal zero. Many times these are the critical points where changes in the function behavior are observed.For the exercise's function \(f(x,y) = 2x^2 + y^2 + 2x - 3y\), the goal is to identify these critical points, both inside the region and along its boundary. These are the steps you should follow:

• Calculate the partial derivatives of the function.
• Set these derivatives to zero to find potential critical points within the interior.
• If necessary due to a constraint, use Lagrange multipliers to find critical points on the boundary.

In our provided example, only boundary points satisfied the condition of being extreme points, due to the inner critical point residing outside the allowed region.

Partial Derivatives

Partial derivatives are foundational tools in calculus when dealing with functions of multiple variables. They measure how a function changes with respect to one variable, while keeping other variables constant. In simpler terms, they help us understand the slope and direction in which a function is changing.To find the partial derivatives of the function \(f(x, y) = 2x^2 + y^2 + 2x - 3y\), you would differentiate with respect to \(x\) and \(y\) separately:

• The partial derivative with respect to \(x\) is found by differentiating the function treating \(y\) as a constant. This gives \(\frac{\partial f}{\partial x} = 4x + 2\).
• The partial derivative with respect to \(y\) involves treating \(x\) as constant, resulting in \(\frac{\partial f}{\partial y} = 2y - 3\).

These derivatives are crucial as they set the stage to find critical points by setting them to zero, which was carried out in the process of solving for extreme points in this exercise.