Problem 21 - 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. $$ f(x, y)=\sin x+\sin y $$

Step-by-Step Solution

Answer

The critical points are \( (x, y) = (\frac{\pi}{2}+k\pi, \frac{\pi}{2}+l\pi) \); they alternate between relative minima and maxima depending on \( k \) and \( l \).

Step 1: Find the partial derivatives

To locate the critical points of the function, we first find the partial derivatives with respect to both variables. For the function \( f(x, y) = \sin x + \sin y \), the partial derivatives are: \[ f_x = \frac{\partial f}{\partial x} = \cos x \] and \[ f_y = \frac{\partial f}{\partial y} = \cos y. \]

Step 2: Set partial derivatives to zero

Set both partial derivatives equal to zero to find critical points:\[ \cos x = 0 \quad \text{and} \quad \cos y = 0. \] This occurs when:\[ x = \frac{\pi}{2} + k\pi \quad \text{and} \quad y = \frac{\pi}{2} + k\pi, \] where \( k \) is any integer.

Step 3: Determine type of critical points using second derivatives

Now we apply the second derivative test. First, compute the second derivatives:\[ f_{xx} = -\sin x, \quad f_{yy} = -\sin y, \quad f_{xy} = 0. \]The Hessian determinant is:\[ H = f_{xx}f_{yy} - (f_{xy})^2 = (-\sin x)(-\sin y) - 0 = \sin x \sin y. \]

Step 4: Analyze sign of the Hessian determinant

For each critical point, substitute the values of \( x \) and \( y \) found in Step 2 into the Hessian:- If \( x = \frac{\pi}{2} + 2m\pi \) and \( y = \frac{\pi}{2} + 2n\pi \) (where \( m, n \) are integers), then \( \sin x = 1 \) and \( \sin y = 1 \). Thus, \( H > 0 \), and these points are relative minima.- If \( x = \frac{\pi}{2} + (2m+1)\pi \) and \( y = \frac{\pi}{2} + (2n+1)\pi \), then \( \sin x = -1 \) and \( \sin y = -1 \). Thus, \( H > 0 \), and these points are relative maxima.- If one partial derivative evaluates to a positive sine while the other evaluates to a negative sine or vice versa, \( H < 0 \) indicating a saddle point. However, for this function, such mixed combinations never occur.

Key Concepts

Partial Derivatives

When dealing with multivariable functions, finding critical points is essential to understanding the behavior of the function. To begin, we use the concept of partial derivatives.
A partial derivative of a function with respect to a variable measures how the function changes when that particular variable is altered, while holding the other variables constant.
For the function \( f(x, y) = \sin x + \sin y \), the partial derivative with respect to \( x \) is \( \frac{\partial f}{\partial x} = \cos x \) and with respect to \( y \) is \( \frac{\partial f}{\partial y} = \cos y \).
By setting these partial derivatives to zero, **\( \cos x = 0 \)** and **\( \cos y = 0 \)**, we identify critical points which are potential candidates for relative maxima, minima, or saddle points.

Second Derivative Test

Once critical points are found, we use the second derivative test to classify them. This involves calculating the second partial derivatives of the function:

• The second partial derivative with respect to \( x \) is \( f_{xx} = -\sin x \).
• The second partial derivative with respect to \( y \) is \( f_{yy} = -\sin y \).
• The mixed partial derivative \( f_{xy} = 0 \).

The second derivative test uses these values in analyzing the behavior of the function near the critical points to determine whether we have a relative maximum, a relative minimum, or a saddle point.

Hessian Determinant

The Hessian determinant is a crucial part of determining the nature of critical points in functions of two variables. It is calculated as:
\[ H = f_{xx}f_{yy} - (f_{xy})^2 \]
For \( f(x, y) = \sin x + \sin y \), this becomes \( H = (-\sin x)(-\sin y) - 0 = \sin x \sin y \).
The Hessian determinant evaluates the concavity around the critical points.

• If \( H > 0 \), the critical point could be a relative maximum or minimum.
• If \( H < 0 \), the function exhibits a saddle point at the critical location.

It allows us to further refine our understanding of the critical points after identifying them.

Relative Maximum

For a critical point to be classified as a relative maximum, both \( f_{xx} \) and \( f_{yy} \) must be negative, indicating that the function curves downwards in every direction.
The Hessian determinant must also be positive at the critical point. In the given function, if \( x = \frac{\pi}{2} + (2m+1)\pi \) and \( y = \frac{\pi}{2} + (2n+1)\pi \), both \( \sin x \) and \( \sin y \) are \(-1\), thus\( H > 0 \). These critical points depict tops of peaks in the function surface, confirming they are relative maxima.

Relative Minimum

Conversely, a relative minimum occurs when the function reaches a low point compared to surrounding values.
In this case, both \( f_{xx} \) and \( f_{yy} \) are positive, creating an upwards curve.
When substituting \( x = \frac{\pi}{2} + 2m\pi \) and \( y = \frac{\pi}{2} + 2n\pi \), we find \( \sin x = 1 \) and \( \sin y = 1 \), leading to a Hessian determinant that is positively valued \( H > 0 \). These configurations ensure the surface is cupped upwards, verifying the presence of relative minima at these points.