Problem 52 - Calculus Early Transcendentals

Question

If possible, find the absolute maximum and minimum values of the following functions on the set \(R\). $$f(x, y)=x+3 y ; R=\\{(x, y):|x|<1,|y|<2\\}$$

Step-by-Step Solution

Answer

Answer: The absolute maximum value is 7 and occurs at points (1, 2) and (-1, 2). The absolute minimum value is -7 and occurs at points (1, -2) and (-1, -2).

Step 1: Identifying Critical Points in the Interior

To find the critical points in the interior of \(R\), we need to calculate the partial derivatives of \(f(x, y)\) with respect to x and y and set them equal to 0. $$\frac{\partial f}{\partial x} = 1$$ $$\frac{\partial f}{\partial y} = 3$$. Since neither of these partial derivatives equates to 0, there are no critical points in the interior of the region \(R\).

Step 2: Evaluating the Function on the Boundary

Now that we have determined that there are no critical points in the interior, we need to check the values of the function along the boundary of \(R\). The boundary can be divided into four segments: 1. The vertical edges (left and right) where \(x = -1\) and \(x = 1\), respectively. 2. The horizontal edges (bottom and top) where \(y=-2\) and \(y=2\), respectively. Let's calculate the function value along each segment:

Step 3: Segment 1: Vertical Edge (x=-1)

We have \(x = -1\), then the function becomes \(f(-1, y) = -1 + 3y\). The maximum and minimum values of this segment are when \(y\) attains its largest and smallest values in the set \(R\), and we have: $$f(-1, -2) = 5$$ $$f(-1, 2) = -7$$

Step 4: Segment 2: Vertical Edge (x=1)

We have \(x = 1\), then the function becomes \(f(1, y) = 1 + 3y\). Similarly, the function attains the maximum and minimum values when \(y\) attains its largest and smallest values in the set \(R\): $$f(1, -2) = -5$$ $$f(1, 2) = 7$$

Step 5: Segment 3: Horizontal Edge (y=-2)

We have \(y = -2\), then the function becomes \(f(x, -2) = x - 6\). The maximum and minimum values of this segment are when \(x\) attains its largest and smallest values in the set \(R\), and we have: $$f(-1, -2) = 5$$ $$f(1, -2) = -5$$

Step 6: Segment 4: Horizontal Edge (y=2)

We have \(y = 2\), then the function becomes \(f(x, 2) = x + 6\). Similarly, the function attains the maximum and minimum values when \(x\) attains its largest and smallest values in the set \(R\): $$f(-1, 2) = -7$$ $$f(1, 2) = 7$$

Step 3: Comparing the Function Values

Now that we have evaluated the function on all boundary segments, let's compare the values to identify the absolute maximum and minimum values: Absolute Maximum: \(7\) at \((1,2)\) and \((-1,2)\) Absolute Minimum: \(-7\) at \((1,-2)\) and \((-1,-2)\)

Key Concepts

Critical Points

When finding the absolute maximum and minimum values of a function, identifying critical points is crucial. Critical points occur where the derivative of the function is zero or undefined. This process is essential to determine points where the function's slope changes, indicating potential maxima, minima, or saddle points.

For functions with several variables, such as in our case with variables \( x \) and \( y \), we need to consider partial derivatives. Partial derivatives measure how a function changes as one variable changes, while all other variables are held constant.

In the problem, we used partial derivatives to find potential critical points. By setting the partial derivatives \(\frac{\partial f}{\partial x} = 1\) and \(\frac{\partial f}{\partial y} = 3\) equal to zero, we attempted to find where the function might achieve a local maximum or minimum.

Since neither of these equations equals zero, there are no critical points within the region \(R\). This outcome directs our search for maxima or minima to the boundaries of the region.

Partial Derivatives

Partial derivatives are a fundamental concept in multivariable calculus, playing a vital role in analyzing functions dependent on more than one variable. They are determined by differentiating the function with respect to one variable, treating all other variables as constants.

For the function \( f(x, y) = x + 3y \), we find the partial derivatives:

• \( \frac{\partial f}{\partial x} = 1 \) - This tells us that the function increases at a constant rate of 1 with respect to \( x \).
• \( \frac{\partial f}{\partial y} = 3 \) - This indicates the function increases at a constant rate of 3 with respect to \( y \).

The constant rates of these partial derivatives highlight that the function is linear, showing no potential changes in slope that might indicate a critical point—meaning no interior maxima or minima within the specified region \( R \).

Understanding these derivatives helps to visualize how the function behaves across different values of \( x \) and \( y \), which is essential for effectively evaluating functions on more complex domains.

Boundary Segment Evaluation

When no critical points occur within a region, the next step involves evaluating the function on the region's boundaries to find absolute extrema. The boundary evaluation process involves checking how the function behaves along the edges of the domain.

In the context of this exercise, the region \( R \) is rectangular, described by \(|x| < 1\) and \(|y| < 2\). This makes the boundary consist of four line segments:

• Vertical edges where \(x = -1\) and \(x = 1\).
• Horizontal edges where \(y = -2\) and \(y = 2\).

For each boundary:- **Vertical edges**, substitute \( x \) with the fixed values to simplify to \( f(x, y) = -1 + 3y \) or \( f(x, y) = 1 + 3y \). Evaluate these expressions as \( y \) varies within its limits.- **Horizontal edges**, substitute \( y \) with the fixed values to simplify to \( f(x, y) = x - 6 \) or \( f(x, y) = x + 6 \). Evaluate as \( x \) varies.
Identifying maximum and minimum function values along these boundaries yields absolute extrema for the specified region. Comparing these values reveals that the absolute maximum is 7, and the absolute minimum is -7 within the region's boundaries, confirming no interior points reach these extremes.