Question

Find the absolute extrema of the function over the region \(R .\) (In each case, \(R\) contains the boundaries.) Use a computer algebra system to confirm your results. $$ \begin{array}{l} f(x, y)=\frac{4 x y}{\left(x^{2}+1\right)\left(y^{2}+1\right)} \\ R=\left\\{(x, y): x \geq 0, y \geq 0, x^{2}+y^{2} \leq 1\right\\} \end{array} $$

Short answer

The short answer to this exercise will be a pair of points representing the absolute maximum and minimum of the function over the region \(R\). The actual values will depend on the specific critical and boundary points found when working through the problem.

Step-by-step solution

Understand the region R

The region \(R\) is defined by three inequalities: \(x \geq 0, y \geq 0, x^{2}+y^{2} \leq 1\). Geometrically, the inequalities \(x \geq 0, y \geq 0\) mean that \(x\) and \(y\) are both non-negative, so we're only concerned with the first quadrant. Further, the inequality \(x^{2}+y^{2} \leq 1\) describes a unit circle, including the interior. So, we know that the region in question is a quarter of a unit disc in the first quadrant.

Find the critical points

Critical points happen where the first partial derivatives of the function are both zero, or if they're undefined. Take the partial derivative of the function \(f(x, y)\) with respect to both \(x\) and \(y\) and set them equal to zero. Solve these equations for \(x\) and \(y\) respectively. Keep in mind that \(x\) and \(y\) are non-negative. Check if these values fall in the region \(R\). These are our critical points inside the region \(R\).

Evaluate the function on the boundary

The boundary of \(R\) in this case is the arc of the unit circle in the first quadrant. To find the extrema along this boundary, we can parameterize it using \(x=cos(t), y=sin(t)\) for \(0 <= t <= \pi/2\). Substitute these into the function and find any critical points (maximums and minimums). Then compute the function's value at these points as well as at the endpoints \(t=0\) and \(t=\pi/2\).

Compare maximums and minimums

Now, compare the function's values at all the critical points and boundary points. The absolute maximum will be the maximum of these values and the absolute minimum will be the minimum of these values.

Key concepts

Partial Derivatives

Partial derivatives are a cornerstone concept in calculus, particularly in the study of functions with several variables. They measure how a function changes as only one of the variables is varied, holding the others constant. For instance, with a function like \( f(x, y) = \frac{4xy}{(x^2+1)(y^2+1)} \), you can find how the function changes in the direction of \( x \) by taking the partial derivative with respect to \( x \). This is often written as \( \frac{\partial f}{\partial x} \).

Computing the partial derivative involves using the same principles as differentiation with respect to a single variable, only treating other variables as constants. In our exercise, we calculate both \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \) to determine how \( f \) changes in both the \( x \) and \( y \) directions independently.

How to Calculate

For \( f(x, y) \), let's break down the computation of one of the partial derivatives:

• To find \( \frac{\partial f}{\partial x} \), look at the function \( f(x, y) \) and differentiate it with respect to \( x \), treating \( y \) as a constant.
• For \( \frac{\partial f}{\partial y} \), do the opposite; differentiate \( f(x, y) \) with respect to \( y \), considering \( x \) constant.

The goal in our exercise is to use these partial derivatives to find critical points where the function might achieve its extrema within the given region \( R \).

Critical Points

Critical points are where the function's behavior could change, potentially indicating the presence of an extremum—either a maximum or minimum—or a saddle point. These points occur where the first partial derivatives equal zero or do not exist.

To find critical points in our exercise, we solve the system of equations created when we set each of the partial derivatives calculated in the previous step to zero. The solutions we find must then be checked to ensure they fall within our designated region \( R \).

Characterizing Critical Points

Once the critical points are obtained, they must be analyzed to understand the function's behavior. In multidimensional calculus, this often involves evaluating the second partial derivatives and using a test like the second derivative test to characterize each point. However, in our particular exercise, we evaluate these points directly since the candidates for extrema include the boundary of the region, which requires a different type of analysis announced in this case by evaluating the function along the boundary.

Optimization in Calculus

Optimization is the process of finding the maximum or minimum values of a function, given a set of constraints. It is a vital application of calculus in many fields, from economics to engineering.

In the context of our exercise, we are tasked with finding the absolute extrema of a given function over a specific region. This involves finding both the highest and lowest values that the function can take. After determining the critical points and boundary points where these extrema could occur, we must compare the function's values at these points.

Optimization Techniques

We use a combination of optimization techniques to solve our problem:

• Finding critical points by setting the partial derivatives equal to zero.
• Evaluating the function's behavior at the boundaries.
• Comparing function values at the identified points to determine the absolute extrema.

This comprehensive approach ensures we consider both the internal behavior of the function and its behavior at the edges of the constraints—both are necessary to find the absolute extrema in the region \( R \).