Problem 26 - Calculus with Concepts in Calculus

Question

Find the extreme values of \(f\) on \(R\). $$ \begin{aligned} &f(x, y)=y e^{-x} ; R \text { is the rectangular region with vertices }\\\ &(0,0),(\ln 2,0),(\ln 2,3),(0,3) \end{aligned} $$

Step-by-Step Solution

Answer

Max is 3 at (0,3); min is 0 at multiple points.

Step 1: Understand the Problem

We need to find the extreme values of the function \[ f(x, y) = y e^{-x} \] within a rectangular region. The vertices of the region are: \( (0,0), (\ln 2, 0), (\ln 2, 3), (0, 3) \). This means we are looking for maxima and minima of \( f(x, y) \) within and along the boundaries of this rectangle.

Step 2: Find Critical Points Inside the Region

To find critical points of \( f(x, y) \), compute the partial derivatives \( f_x \) and \( f_y \), set them to zero, and solve:\[ f_x = -y e^{-x} = 0 \text{ implies } y = 0. \]\[ f_y = e^{-x} = 0 \text{ implies no solution since } e^{-x} > 0 \text{ for all x.} \]Thus, there are no critical points inside the region where \( f_x \) and \( f_y \) both are zero.

Step 3: Evaluate Function on Boundary

We evaluate \( f \) on the boundary of the rectangle.1. **For y = 0**: \( f(x, 0) = 0 \)2. **For y = 3**: \( f(x, 3) = 3e^{-x} \)3. **For x = 0**: \( f(0, y) = y \)4. **For x = \ln 2**: \( f(\ln 2, y) = y e^{-(\ln 2)} = \frac{y}{2} \)

Step 4: Calculate Extreme Values on the Boundary

Evaluate \( f \) on each boundary segment's end points:- **y=0, x varies 0 to \ln 2**: \( f(0, 0) = 0, f(\ln 2, 0) = 0 \).- **y=3, x varies 0 to \ln 2**: \( f(0, 3) = 3, f(\ln 2, 3) = \frac{3}{2} \).- **x=0, y varies 0 to 3**: \( f(0, 0) = 0, f(0, 3) = 3 \).- **x=\ln 2, y varies 0 to 3**: \( f(\ln 2, 0) = 0, f(\ln 2, 3) = \frac{3}{2} \).

Step 5: Identify the Extreme Values

The maximum value of \( f(x, y) \) is 3 (occurs at \( (0,3) \)), and the minimum value is 0 (occurs at points \( (0,0), (\ln 2,0), \) and \( (\ln 2,3) \)).

Key Concepts

Partial Derivatives

When dealing with functions of two or more variables, partial derivatives are an essential tool. They help us understand how the function changes as we vary one variable at a time, while keeping others constant. For the function \[f(x, y) = y e^{-x}\]we find the partial derivatives \( f_x \) and \( f_y \) to analyze how the function changes with respect to \( x \) and \( y \), respectively.

• \( f_x \) is the derivative of \( f \) with respect to \( x \), treating \( y \) as constant. For our function, \( f_x = -y e^{-x} \).
• \( f_y \) is the derivative of \( f \) with respect to \( y \), treating \( x \) as constant. Here, \( f_y = e^{-x} \).

To find critical points, we set these derivatives equal to zero and solve. This determines where the rate of change in these directions is zero, a potential location for extreme values.

Critical Points

Critical points of a function are points where the partial derivatives are zero or undefined. At these points, the function does not increase or decrease in the vicinity, indicating potential locations for maxima or minima.
For \( f(x, y) = y e^{-x} \):

• Setting \( f_x = -y e^{-x} = 0 \) leads us to conclude \( y = 0 \).
• Setting \( f_y = e^{-x} = 0 \) reveals no solutions because \( e^{-x} \) is always positive for real \( x \).

Hence, there are no interior critical points where both the partial derivatives are zero. We then move to evaluating the function on the boundary of the region.

Boundary Evaluation

Evaluating the function on the boundary is crucial when searching for extreme values over a closed region. Since critical points inside \( R \) yielded no results, we analyze the behavior of \( f \) along the boundary of the rectangular region.
Consider each segment:

• For \( y = 0 \), the function simplifies to \( f(x, 0) = 0 \) for all \( x \).
• For \( y = 3 \), the function becomes \( f(x, 3) = 3e^{-x} \).
• For \( x = 0 \), \( f(0, y) = y \) reflects a linear increase in \( y \).
• For \( x = \ln 2 \), reducing the function yields \( f(\ln 2, y) = \frac{y}{2} \).

This analysis helps us identify how values change along each boundary, setting the stage to find actual extreme values.

Rectangular Region

We are working within a defined rectangular region described by vertices at \( (0,0), (\ln 2,0), (\ln 2,3), (0,3) \). This region sets the bounds within which we inspect for extreme values.
The boundaries we evaluate are essentially lines between these vertices. Each segment of the rectangle offers different function behaviors:

• An understanding of how \( f \) behaves along \( y = 0 \) and \( y = 3 \) reveals horizontal changes.
• Vertical changes are noted on segments \( x = 0 \) and \( x = \ln 2 \).

By examining each segment, including endpoints, we determine \( f(0,3) = 3 \) at \( (0,3) \) as the maximum value, while \( f \) reaches a minimum of 0 at any point where \( x = \ln 2 \) or \( y = 0 \). Understanding these boundaries is key to fully solving the problem.