Problem 20 - Calculus with Concepts in Calculus

Question

Find all critical points. Determine whether each critical point yields a relative maximum value, a relative minimum value, or a saddle point. $$ f(x, y)=\sin x+\sin y \text { for } 0

Step-by-Step Solution

Answer

There are no critical points within the specified domain.

Step 1: Find the First Partial Derivatives

Calculate the first partial derivatives of the function with respect to both variables. The partial derivative with respect to \( x \) is \( f_x(x, y) = \cos x \) and with respect to \( y \) is \( f_y(x, y) = \cos y \).

Step 2: Set the First Partial Derivatives to Zero

To find the critical points, set the first partial derivatives equal to zero. Thus, \( \cos x = 0 \) and \( \cos y = 0 \). This implies that \( x = \frac{\pi}{2} \) and \( y = \frac{\pi}{2} \) are the solutions.

Step 3: Identify the Critical Points Within the Given Domain

Given the domain \( 0 < x < \frac{\pi}{2} \) and \( 0 < y < \frac{\pi}{2} \), the solution \((\frac{\pi}{2}, \frac{\pi}{2})\) is on the boundary and thus not within the interior of the given domain. There are no critical points within the given interior domain.

Step 4: Analyze the Second Derivatives for Classification

Since there are no critical points within the domain of interest, we do not need to evaluate the second derivatives or the Hessian matrix for classification. Therefore, we cannot determine any relative maxima, minima, or saddle points within \( 0 < x < \frac{\pi}{2} \) and \( 0 < y < \frac{\pi}{2} \).

Key Concepts

Partial Derivative

Partial derivatives help us understand how a function changes as each variable changes independently. For functions like \( f(x, y) = \sin x + \sin y \), we find the change with respect to \( x \) while keeping \( y \) constant, and vice versa.

• The partial derivative with respect to \( x \), denoted by \( f_x(x, y) \), measures how \( f \) changes when \( x \) varies, giving us \( \cos x \).
• Similarly, the partial derivative with respect to \( y \), \( f_y(x, y) \), shows the change with \( y \), providing \( \cos y \).

By setting these derivatives to zero, we find stationary points. These are places where the function doesn't have a steep slope in any direction, which can potentially be where maxima, minima, or saddle points occur.Finding the partial derivatives and setting them to zero is crucial because critical points typically occur at these stationary positions.

Relative Maximum

A relative maximum point is where the function has a peak within a small neighborhood. At this point, the function value is higher than all nearby points. We usually use the second derivative test to confirm a relative maximum:

• If the second partial derivatives are negative, the critical point is a relative maximum.
• The usual test involves evaluating the Hessian matrix, but we need critical points to apply it.

In our exercise, we couldn't identify any critical points within the domain \( 0 < x < \frac{\pi}{2}, 0 < y < \frac{\pi}{2} \). Therefore, we couldn't determine any relative maxima within these constraints.

Relative Minimum

A relative minimum occurs at a point where the function is lower than all nearby points. It's like a little "valley" in the graph. To find it, we can use the second derivative test:

• If the second partial derivatives are positive, the critical point is a relative minimum.
• We leverage the Hessian matrix for this verification in typical scenarios when critical points exist.

In our scenario, since no critical points were found within \( 0 < x < \frac{\pi}{2} \) and \( 0 < y < \frac{\pi}{2} \), we can't locate any relative minima in the given domain.

Saddle Point

A saddle point is special because it's neither a maximum nor a minimum. Instead, the function moves upwards in one direction and downwards in another, much like a saddle.To identify a saddle point, we usually check:

• The second derivative test, where if the Hessian matrix determinant is negative, it indicates a saddle point.

However, since our function did not have any critical points in the specified domain of \( 0 < x < \frac{\pi}{2} \) and \( 0 < y < \frac{\pi}{2} \), we couldn't classify any saddle points either.In functions like our example, where critical points may border the defined area, sometimes examining boundary trends or extending domains may offer insights, but under given limits, no saddle points were detectable.