Problem 54 - Calculus Early Transcendentals

Question

If possible, find the absolute maximum and minimum values of the following functions on the set \(R\). $$f(x, y)=x^{2}-y^{2} ; R=\\{(x, y) ;|x|<1,|y|<1\\}$$

Step-by-Step Solution

Answer

Answer: The absolute maximum value is 1 and the absolute minimum value is -1.

Step 1: Find the partial derivatives

Calculate the partial derivatives of the function with respect to \(x\) and \(y\): $$f_x = \frac{\partial f}{\partial x} = 2x.$$ $$f_y = \frac{\partial f}{\partial y} = -2y.$$

Step 2: Find the critical points

Set the partial derivatives equal to zero and solve for \(x\) and \(y\): $$2x = 0 \Rightarrow x = 0,$$ $$-2y = 0 \Rightarrow y = 0.$$ Thus, we have one critical point at \((0, 0)\).

Step 3: Use the second partial derivative test

Find the second partial derivatives: $$f_{xx} = \frac{\partial^2 f}{\partial x^2} = 2,$$ $$f_{yy} = \frac{\partial^2 f}{\partial y^2} = -2,$$ $$f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = 0.$$ Calculate the discriminant \(D = f_{xx}f_{yy} - f_{xy}^2 = (2)(-2) - 0^2 = -4\). Since \(D < 0\), the second derivative test is inconclusive. We will need to check the boundary of the region.

Step 4: Check the boundary values

To find the maximum and minimum values, we need to check the boundary of the set \(R = \{(x, y)\:|\: |x| < 1, |y| < 1\}\). The boundary of R consists of four lines: \(x = \pm 1\) and \(y = \pm 1\). Calculate the function values at these boundaries: For \(x = 1\), \(f(1, y) = 1-y^2\). The minimum will obviously occur when \(y^2\) has its maximum value of 1, which is achieved at \(y = \pm 1\). The maximum will occur when \(y^2\) has its minimum value of 0, which occurs at \(y = 0\). Therefore, we have \(f(1, -1)=f(1, 1)=0\) and \(f(1,0)=1\). For \(x=-1\), \(f(-1,y)=1-y^2\). The function values are the same as for \(x=1\); hence, we have \(f(-1,-1)=f(-1,1)=0\) and \(f(-1,0)=1\). For \(y=1\), \(f(x,1)=x^2-1\). The minimum occurs when \(x^2\) has its maximum value of 1, which is achieved at \(x = \pm 1\). The maximum occurs when \(x^2\) has its minimum value of 0, which occurs at \(x = 0\). Therefore, we have \(f(-1,1)=f(1,1)=0\) and \(f(0,1)=-1\). For \(y=-1\), \(f(x,-1)=x^2-1\). The function values are the same as for \(y=1\); hence, we have \(f(-1,-1)=f(1,-1)=0\) and \(f(0,-1)=-1\). Comparing all the values, we find that the absolute maximum is \(f(1,0)=f(-1,0)=1\) and the absolute minimum is \(f(0,1)=f(0,-1)=-1\).

Key Concepts

Partial Derivatives

Partial derivatives help us find how a function with multiple variables changes as each variable changes, while keeping other variables constant. For the function given, \( f(x, y) = x^2 - y^2 \), the partial derivative with respect to \( x \) is represented as \( f_x = \frac{\partial f}{\partial x} \). To calculate this, treat \( y \) as a constant and differentiate the function with respect to \( x \). The result is \( f_x = 2x \).
Similarly, the partial derivative with respect to \( y \), \( f_y = \frac{\partial f}{\partial y} \), treats \( x \) as a constant, leading to \( f_y = -2y \). These derivatives tell us the slope of the function in each direction and are the first step in identifying the critical points.

Partial derivatives play a pivotal role in multivariable calculus. They allow you to explore the shape and trends of surfaces, especially when solving optimization problems like finding maxima or minima.

Critical Points

Critical points occur where the partial derivatives of a function are zero or undefined. These points are crucial because they are potential locations for relative maxima, minima, or saddle points. To find the critical points for \( f(x, y) = x^2 - y^2 \), set the partial derivatives \( f_x = 2x \) and \( f_y = -2y \) to zero.
This gives rise to the equations:

• \( 2x = 0 \) which means \( x = 0 \)
• \( -2y = 0 \) which means \( y = 0 \)

Therefore, the only critical point is \( (0, 0) \). At this point, the function's behavior needs to be further analyzed to determine the nature of extremum, if any. Understanding critical points is a foundational skill in calculus and aids in graphically interpreting where significant changes in a function's surface occur.

Second Partial Derivative Test

The second partial derivative test helps classify critical points into local maxima, minima, or saddle points. For function \( f(x, y) \), calculate the second partial derivatives:

• \( f_{xx} = \frac{\partial^2 f}{\partial x^2} = 2 \)
• \( f_{yy} = \frac{\partial^2 f}{\partial y^2} = -2 \)
• \( f_{xy} = f_{yx} = \frac{\partial^2 f}{\partial x \partial y} = 0 \)

Next, compute the discriminant, \( D \), where \( D = f_{xx}f_{yy} - f_{xy}^2 \). Substituting the values, we find \( D = (2)(-2) - 0^2 = -4 \).
A negative discriminant \( (D < 0) \) indicates a saddle point at \( (0, 0) \). In this case, the test is inconclusive for finding absolute extrema, thus guiding us to further check boundary values. Understanding the second derivative test is essential for diagnosing the behavior of multivariable functions at their critical points.

Boundary Values

When the second derivative test is inconclusive, it's important to assess the boundary values of the region where the function is defined. For the problem's domain, \( R = \{(x, y) \: | \: |x| < 1, |y| < 1\} \), boundaries occur as \( x = \pm1 \) and \( y = \pm1 \).

• For \( x = 1 \) and \( x = -1 \), evaluate \( f(x, y) = 1 - y^2 \). The value is maximized to 1 when \( y = 0 \), and zero when \( y = \pm 1 \).
• Similarly, for \( y = 1 \) and \( y = -1 \), evaluate \( f(x, y) = x^2 - 1 \). The value is minimized to -1 when \( x^2 = 0 \) and zero when \( x = \pm1 \).

After examining the boundaries, the absolute maximum value of the function is 1 at points \((1,0)\) and \((-1,0)\), while the absolute minimum value is -1 occurring at \((0,1)\) and \((0,-1)\). Checking boundary values ensures that potential extremum are not missed when working with closed regions.