Problem 24 - Calculus Early Transcendentals

Question

Use a graph and/or level curves to estimate the local maximum and minimum values and saddle point(s) of the function. Then use calculus to find these values precisely. $$\begin{array}{l}{f(x, y)=\sin x+\sin y+\cos (x+y)} \\ {0 \leqslant x \leqslant \pi / 4,0 \leqslant y \leqslant \pi / 4}\end{array}$$

Step-by-Step Solution

Answer

The local maximum is \( f(\pi/4, \pi/4) = \sqrt{2} \). No local minima or saddle points are present in the domain.

Step 1: Understand the Function

We are given the function \( f(x, y) = \sin x + \sin y + \cos (x+y) \) and boundaries \( 0 \leq x \leq \pi/4 \) and \( 0 \leq y \leq \pi/4 \). This function is a combination of sine and cosine functions.

Step 2: Graphing the Function

Plot the function \( f(x, y) \) within the specified domain to visually identify the maximum, minimum, and saddle points. These are regions on the graph where the function values appear to reach highs or lows or where the surface changes direction.

Step 3: Calculate the Partial Derivatives

Calculate the partial derivatives to find the critical points. The partial derivatives are: \(\frac{\partial f}{\partial x} = \cos x - \sin(x+y),\quad \frac{\partial f}{\partial y} = \cos y - \sin(x+y). \)

Step 4: Set the Partial Derivatives to Zero

Find the critical points by solving \( \frac{\partial f}{\partial x} = 0 \) and \( \frac{\partial f}{\partial y} = 0 \): \(\cos x = \sin(x+y),\quad \cos y = \sin(x+y). \) Solve these equations simultaneously.

Step 5: Solve the System of Equations

To solve the system of equations \( \cos x = \cos y \), and \( \cos x = \sin(x+y) \), conclude that one simple critical point is \( x = y = \pi/4 \).

Step 6: Evaluate the Function at Critical Points

Substitute the critical point \( x = y = \pi/4 \) into the function to evaluate \( f(x,y) \). For \( x = \pi/4, y = \pi/4 \), \( f(x, y) = \sin(\pi/4) + \sin(\pi/4) + \cos(\pi/2) = \sqrt{2}/2 + \sqrt{2}/2 + 0 = \sqrt{2}. \)

Step 7: Use Second Partial Derivative Test

To verify the nature of the critical points, calculate the second derivatives: \(\frac{\partial^2 f}{\partial x^2} = -\sin x - \cos(x+y),\quad \frac{\partial^2 f}{\partial y^2} = -\sin y - \cos(x+y),\frac{\partial^2 f}{\partial x \partial y} = -\cos(x+y)\)Evaluate these to determine the type of critical points using the discriminant \( D = \frac{\partial^2 f}{\partial x^2} \frac{\partial^2 f}{\partial y^2} - \left( \frac{\partial^2 f}{\partial x \partial y} \right)^2. \) If \( D > 0 \) and \( \frac{\partial^2 f}{\partial x^2} < 0 \), it's a max; if \( \frac{\partial^2 f}{\partial x^2} > 0 \), it's a min. If \( D < 0 \), it's a saddle point.

Step 8: Final Evaluation of the Critical Points

After computing, if \( D > 0 \) and \( \frac{\partial^2 f}{\partial x^2} < 0 \), confirm \( x = y = \pi/4 \) is a local maximum with value \( \sqrt{2} \). Check other points with boundary values to ensure no other maxima or minima within the given domain.

Key Concepts

Local Extrema

Local extrema refer to points in a function's domain where the function takes on a maximum or minimum value in a small neighborhood around those points. In multivariable calculus, local extrema can be thought of as the peaks (local maxima) and valleys (local minima) of a surface, while saddle points are areas where the function does not have a direct peak or valley but instead resembles a saddle.

To find local extrema, you need to analyze the regions on the function’s graph where these local maxima, minima, or saddle points occur. Visual inspectin – through graphs or level curves – helps in giving an initial estimate of where these points might be. However, actual verification requires calculus techniques, particularly the use of partial derivatives, to pinpoint these locations precisely.

In the given function, we are primarily interested in finding these extreme points within the specified domain defined by the intervals for x and y.

Partial Derivatives

Partial derivatives help us understand how a function changes as each variable is varied independently, while keeping the other variable constant. In the context of multivariable calculus functions like our example, partial derivatives are crucial for exploring changes in the surface.

The partial derivative with respect to x, denoted as \( \frac{\partial f}{\partial x} \), gives us the rate of change of the function as x changes. Similarly, the partial derivative with respect to y, \( \frac{\partial f}{\partial y} \), focuses on changes along the y direction.

In our specific problem,

• For \( \frac{\partial f}{\partial x} = \cos x - \sin(x+y) \), we understand how the function reacts horizontally.
• For \( \frac{\partial f}{\partial y} = \cos y - \sin(x+y) \), we observe vertical behavior.

Setting these partial derivatives to zero is the key step to finding critical points, as it helps identify where the function stops increasing or decreasing - potential spots for local extrema.

Second Partial Derivative Test

The second partial derivative test is a method used to classify critical points found using the first derivatives. This test helps determine whether these critical points are local minima, local maxima, or saddle points.

For a function \( f(x, y) \), the second partial derivatives involved are:

• \( \frac{\partial^2 f}{\partial x^2} \) - the curvature in the direction of x
• \( \frac{\partial^2 f}{\partial y^2} \) - the curvature in the direction of y
• \( \frac{\partial^2 f}{\partial x \partial y} \) - the mixed derivative indicating how x and y interact

Using these, we calculate the discriminant, \( D = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - \left( \frac{\partial^2 f}{\partial x \partial y} \right)^2 \). The sign and conditions of D tell us:

• If \( D > 0 \) and \( \frac{\partial^2 f}{\partial x^2} < 0 \), we have a local maximum.
• If \( D > 0 \) and \( \frac{\partial^2 f}{\partial x^2} > 0 \), it's a local minimum.
• If \( D < 0 \), it indicates a saddle point.

In our solution, this test helps confirm whether the point \( (\pi/4, \pi/4) \) is indeed a local extremum.

Critical Points

Critical points are the locations on a surface where the partial derivatives are zero. This means there is no rate of change in any direction from the point. Identifying these points is crucial for finding where the function might achieve local maxima, minima, or neither.

The process involves:

• Calculating the partial derivatives of the function.
• Setting these derivatives equal to zero to form equations.
• Solving these equations simultaneously to find specific values of x and y, which are the critical points.

For the function \( f(x, y) = \sin x + \sin y + \cos (x+y) \), we identified and solved:

• \( \cos x = \sin(x+y) \)
• \( \cos y = \sin(x+y) \)

By finding solutions such as \( x = y = \pi/4 \), we ascertain the potential location of local extrema within the given domain, which then undergo further testing for confirmation of their nature.