Problem 44 - Calculus Early Transcendentals: Pearson New International Edition

Question

\(f(x, y)=y /\left(1+x^{2}+y^{2}\right) ;-5 \leq x \leq 5,-5 \leq y \leq 5\); global maximum point and global minimum. Check using calculus.

Step-by-Step Solution

Answer

Global maximum: \( \frac{1}{2} \) at \( (0, 1) \); Global minimum: \( -\frac{1}{2} \) at \( (0, -1) \).

Step 1: Find Partial Derivatives

To find the critical points, we need to find the partial derivatives of the function \( f(x, y) = \frac{y}{1 + x^2 + y^2} \). Let's compute the partial derivatives with respect to \( x \) and \( y \). The partial derivative with respect to \( x \) is \( f_x = -\frac{2xy}{(1+x^2+y^2)^2} \), and the partial derivative with respect to \( y \) is \( f_y = \frac{1 + x^2 - y^2}{(1+x^2+y^2)^2} \).

Step 2: Set Partial Derivatives to Zero

Next, we find the critical points by setting both partial derivatives to zero: \( f_x = 0 \) and \( f_y = 0 \). This gives us \( -\frac{2xy}{(1+x^2+y^2)^2} = 0 \), which implies either \( x = 0 \) or \( y = 0 \). The equation \( \frac{1 + x^2 - y^2}{(1+x^2+y^2)^2} = 0 \) implies \( 1 + x^2 - y^2 = 0 \), so \( y^2 = 1 + x^2 \).

Step 3: Solve the System of Equations

Solving the equations \( x = 0 \) or \( y = 0 \) together with \( y^2 = 1 + x^2 \), we find the potential critical points. For \( x = 0 \), \( y^2 = 1 \), giving \( y = \pm 1 \). For \( y = 0 \), there is no solution satisfying \( y^2 = 1 + x^2 \). Hence, our critical points are \( (0, 1) \) and \( (0, -1) \).

Step 4: Evaluate the Function at Boundary and Critical Points

Evaluate \( f(x, y) \) at the critical points and along the boundary \(-5 \leq x \leq 5, -5 \leq y \leq 5\). At \((0, 1)\), \( f(0, 1) = \frac{1}{2} \); at \((0, -1)\), \( f(0, -1) = -\frac{1}{2} \). On the lines \( y = 5 \) and \( y = -5 \): \( f(x, 5) = \frac{5}{1+x^2+25} \) and \( f(x, -5) = \frac{-5}{1+x^2+25} \). As \( x \to \pm 5 \), these expressions tend towards zero. Similar results hold on \( x = 5 \) and \( x = -5 \).

Step 5: Identify Global Maximum and Minimum

Comparing values found at the critical points and along boundaries, the global maximum occurs at \( (0, 1) \) with value \( \frac{1}{2} \) and the global minimum occurs at \( (0, -1) \) with value \(-\frac{1}{2} \). No other points on the boundaries or within the domain offer larger or smaller values than these.

Key Concepts

Partial Derivatives

Partial derivatives are essential in multivariable calculus, especially when dealing with functions involving more than one variable. They provide insight into how a function changes as each variable is varied independently. For example, with a function like \( f(x, y) = \frac{y}{1 + x^2 + y^2} \), we compute partial derivatives by differentiating with respect to one variable while keeping the others constant.

For the given function, the partial derivative with respect to \( x \) is \( f_x = -\frac{2xy}{(1+x^2+y^2)^2} \) and with respect to \( y \) is \( f_y = \frac{1 + x^2 - y^2}{(1+x^2+y^2)^2} \). These expressions tell us how the function \( f \) changes as \( x \) or \( y \) is varied independently. Understanding these changes is crucial for further analysis, such as identifying critical points.

Critical Points

Critical points occur where the gradient (think of it as the slope in higher dimensions) is zero. For functions of two variables, it means both partial derivatives must be zero at the critical point.

In our example, setting \( f_x = 0 \) gives \(-\frac{2xy}{(1+x^2+y^2)^2} = 0\), implying \( x = 0 \) or \( y = 0 \). Meanwhile, setting \( f_y = 0 \) gives \( y^2 = 1 + x^2 \). Solving these equations together gives us critical points \( (0, 1) \) and \( (0, -1) \).

These points are important; they are potential locations for local maxima, minima, or saddle points.

Global Maximum and Minimum

To find the global maximum and minimum, you consider both the interior of the domain and the boundaries, if applicable. Once critical points are identified, evaluate the function at these points.

For our function, at critical points \( (0, 1) \) and \( (0, -1) \), the function values are \( \frac{1}{2} \) and \( -\frac{1}{2} \), respectively. These are potential candidates for global extrema.

In addition, it's necessary to compare these values with those at the boundary to confirm they are the largest and smallest in the given domain.

Boundary Evaluation

Evaluating the function on the boundaries of the domain ensures that you account for every possible point that might have the highest or lowest function value.

For the given function, evaluate on the boundary lines, such as \( y = 5 \), \( y = -5 \), \( x = 5 \), and \( x = -5 \). Here, the calculations show that as \( x \) approaches these boundaries, the function tends towards zero. This comparison ensures confidence that neither higher maxima nor lower minima exist on the boundaries than those found at critical points.

Such evaluations affirm that the global maximum and minimum values indeed occur at the critical points in the interior of the domain.