Problem 27 - Calculus with Concepts in Calculus

Question

Find the extreme values of \(f\) on \(R\). $$ \begin{aligned} &f(x, y)=2 \sin x+3 \cos y ; R \text { is the square region with }\\\ &\text { vertices }(0,-\pi / 2),(\pi,-\pi / 2),(\pi, \pi / 2),(0, \pi / 2) \text { . } \end{aligned} $$

Step-by-Step Solution

Answer

The maximum value is 5 and the minimum value is 0.

Step 1: Identify Constraints and Function

The function to be analyzed is \( f(x, y) = 2\sin x + 3\cos y \) and it is defined over the region \( R \), which is a square with vertices \((0, -\pi/2), (\pi, -\pi/2), (\pi, \pi/2), (0, \pi/2)\). The problem is to find the maximum and minimum values (extreme values) within this region.

Step 2: Find Critical Points Inside the Region

To find critical points, calculate the partial derivatives \( f_x \) and \( f_y \). Set each derivative to 0 and solve for critical points. \[ f_x = \frac{\partial}{\partial x}(2\sin x + 3\cos y) = 2\cos x \] \[ f_y = \frac{\partial}{\partial y}(2\sin x + 3\cos y) = -3\sin y \] Solving \( f_x = 0 \) gives \( \cos x = 0 \) (\( x = \frac{\pi}{2} \)). Solving \( f_y = 0 \) gives \( \sin y = 0 \) (\( y = 0 \)). Thus, the critical point inside the region is \((x, y) = \left(\frac{\pi}{2}, 0\right)\).

Step 3: Evaluate Function at Critical Points

Plug the critical point into the function to find its value. \[ f\left(\frac{\pi}{2}, 0\right) = 2\sin\left(\frac{\pi}{2}\right) + 3\cos(0) = 2(1) + 3(1) = 5 \]

Step 4: Evaluate Function on Boundaries

Analyze the boundaries (sides of the square) and evaluate the function:- Bottom side \(y = -\frac{\pi}{2}\): \[ f(x, -\frac{\pi}{2}) = 2\sin x + 3\cos(-\frac{\pi}{2}) = 2\sin x \] Maximum and minimum of \(2\sin x\) on \([0, \pi]\) are 2 and 0.- Top side \(y = \frac{\pi}{2}\): \[ f(x, \frac{\pi}{2}) = 2\sin x + 3\cos\left(\frac{\pi}{2}\right) = 2\sin x \] Maximum and minimum of \( 2\sin x \) on \([0, \pi]\) are 2 and 0.- Left side \(x = 0\): \[ f(0, y) = 2\sin(0) + 3\cos y = 3\cos y \] Maximum and minimum of \(3\cos y \) on \([-\frac{\pi}{2}, \frac{\pi}{2}]\) are 3 and 0.- Right side \(x = \pi\): \[ f(\pi, y) = 2\sin(\pi) + 3\cos y = 3\cos y \] Maximum and minimum of \(3\cos y\) on \([-\frac{\pi}{2}, \frac{\pi}{2}]\) are 3 and 0.

Step 5: Compare All Values

Compare all calculated values:- Critical point \( f\left( \frac{\pi}{2}, 0 \right) = 5 \)- Boundary sides: - Bottom and Top boundaries: maximum 2, minimum 0. - Left and Right boundaries: maximum 3, minimum 0.The highest value found is 5 and the lowest value is 0.

Key Concepts

Critical Points

In calculus, critical points of a function are essential for determining extreme values, such as maximum and minimum points. This is because they offer insights into where the slope or rate of change of a function could be zero. For a function of two variables, like in our given scenario, we locate critical points by evaluating partial derivatives and equating them to zero.Here's what happens step-by-step:

• Partial Derivatives: Derivatives are calculated with respect to each variable, treating other variables as constants.
• Setting Equal to Zero: Each partial derivative is set to zero to find critical points.
• Solving: Solutions to these equations give the critical points within the defined region.

For the provided function, \[ f(x, y) = 2 \sin x + 3 \cos y \]we solve:

• Partial with respect to x gives \( f_x = 2\cos x \). Setting \( 2\cos x = 0 \), we find \( x = \frac{\pi}{2} \).
• Partial with respect to y gives \( f_y = -3\sin y \). Setting \( -3\sin y = 0 \), we find \( y = 0 \).

Thus, the critical point in the region is \((\frac{\pi}{2}, 0)\).

Partial Derivatives

Partial derivatives are a cornerstone in multivariable calculus, especially when finding critical points of functions with more than one variable. They allow us to understand the behavior of a function, holding all variables constant except one.In the example exercise, the partial derivatives of the function \( f(x, y) \) are calculated as follows:

• Partial Derivative with respect to x: \[ f_x = \frac{\partial}{\partial x}(2\sin x + 3\cos y) = 2\cos x \]
• Partial Derivative with respect to y: \[ f_y = \frac{\partial}{\partial y}(2\sin x + 3\cos y) = -3\sin y \]

These derivatives tell us how the function changes along the x and y dimensions, respectively. By setting them to zero, we identify positions where the function's slope vanishes, which are potentially extreme (maximum or minimum) points.

Boundary Values

Boundary values are crucial when a function is defined over a specific region. These are values the function might take along the boundary of that region, where the potential for extremes (maxima or minima) can occur.For the function \( f(x, y) = 2\sin x + 3\cos y \) in the square region, we need to evaluate the function around the edges. Each edge or boundary modifies the function to a simpler one-variable form:

• Bottom Edge \( y = -\frac{\pi}{2} \): The function becomes \( f(x, -\frac{\pi}{2}) = 2\sin x \). Here, we evaluate sin function's behavior over \([0, \pi] \).
• Top Edge \( y = \frac{\pi}{2} \): Similarly \( f(x, \frac{\pi}{2}) = 2\sin x \) evaluated over \([0, \pi] \).
• Left Edge \( x = 0 \): Translates to \( f(0, y) = 3\cos y \) across \([-\frac{\pi}{2}, \frac{\pi}{2}] \).
• Right Edge \( x = \pi \): Becomes \( f(\pi, y) = 3\cos y \) over \([-\frac{\pi}{2}, \frac{\pi}{2}] \).

Through this analysis, the function's extreme values along these boundaries are found and compared against each other.

Function Evaluation

Function evaluation is the process of determining the actual numerical outcome of a function given specific inputs or conditions. This concept is vital when assessing the potential extremes discovered from critical and boundary analysis.For our example function \( f(x, y) = 2\sin x + 3\cos y \), evaluation at the calculated critical point and along boundaries involves:

• At Critical Point: We determine \( f\left(\frac{\pi}{2}, 0\right) = 5 \). This gives a direct value at the critical point.
• Along Boundaries: Evaluate simplified forms such as \( 2\sin x \) and \( 3\cos y \) for the specific range of x and y as described in the boundary evaluation section.

Through these evaluations, the highest and lowest function values across the region are found, providing a comprehensive picture of extreme values. This process is essential to ensure all possible extremes are considered, both within the interior of the region and along its boundaries.