Problem 14 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the global maximum value and global minimum value of \(f\) on \(S\) and indicate where each occurs. \(f(x, y)=x^{2}-6 x+y^{2}-8 y+7 ;\) \(S=\left\\{(x, y): x^{2}+y^{2} \leq 1\right\\}\)

Step-by-Step Solution

Answer

The global maximum is 14, occurring at \\(1, 0\\), and the global minimum is -9, occurring at \\(-3/5, 4/5\\).

Step 1: Understand the Problem

Identify the function and the domain. The function is \(f(x, y) = x^2 - 6x + y^2 - 8y + 7\). The region \(S\) is defined as the set of all points \((x, y)\) such that \(x^2 + y^2 \leq 1\). This represents a closed disk centered at the origin with radius 1.

Step 2: Find Critical Points in the Interior of Region S

To find critical points, calculate the partial derivatives of \(f\) and set them to zero. The partial derivative with respect to \(x\) is \(f_x = 2x - 6\) and with respect to \(y\) it is \(f_y = 2y - 8\). Set \(f_x = 0\) to find \(x = 3\) and \(f_y = 0\) to find \(y = 4\). However, \((3,4)\) is outside the region \(S\). Thus, there are no critical points within the disk.

Step 3: Examine the Boundary of Region S

The boundary is defined by \(x^2 + y^2 = 1\). Substitute \(y^2 = 1 - x^2\) into \(f(x, y)\). This results in a new function: \(f_{boundary}(x) = x^2 - 6x + (1 - x^2) - 8y + 7\). Simplify: \(f(x, y) = 1 + x^2 - 6x - x^2 + 7\). Now calculate values on the boundary.

Step 4: Parametrize Boundary and Calculate Values

To simplify evaluating the function on the boundary, use polar coordinates where \(x = \cos\theta\) and \(y = \sin\theta\). Substitute these into \(f\): \(f( heta) = \cos^2\theta - 6\cos\theta + \sin^2\theta - 8\sin\theta + 7\). Simplify \(f\) by noting that \(\cos^2\theta + \sin^2\theta = 1\), so \(f(\theta) = 1 - 6\cos\theta - 8\sin\theta + 7\).

Step 5: Maximize and Minimize f Along the Boundary

To find the extrema, compute the derivative of \(f(\theta)\), set it to zero and solve. The derivative is \(f'(\theta) = 6\sin\theta - 8\cos\theta\). Setting \(f'(\theta) = 0\) gives \(\tan\theta = \frac{8}{6}\). Solving gives critical angles. Evaluate \(f\) at these angles and key points like \(\theta = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}\).

Step 6: Evaluate Function at Key Points

Evaluate \(f(\theta)\) at \(\theta = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}\), as well as angles obtained from solving \(\tan\theta = \frac{4}{3}\). Compute the corresponding \(x, y\) and substitute these back into \(f(x, y)\). Compare values to determine the maximum and minimum on boundary.

Key Concepts

Partial Derivatives

Calculating partial derivatives is vital in understanding how a multivariable function changes with respect to each independent variable. In this problem, the function is given as \( f(x,y) = x^2 - 6x + y^2 - 8y + 7 \). We find the partial derivative with respect to \( x \) by keeping \( y \) constant, resulting in \( f_x = 2x - 6 \). Similarly, we find the partial with respect to \( y \), \( f_y = 2y - 8 \).

• Setting \( f_x \) to zero helps us find points where the slope of the function (in the \( x \) direction) is zero, indicating potential critical points.
• Setting \( f_y = 0 \) finds points where the slope of the function (in the \( y \) direction) is zero.

In optimization problems, examining where both partial derivatives are zero helps identify critical points inside the region, if any exist. Here, solving \( 2x - 6 = 0 \) gives \( x = 3 \) and \( 2y - 8 = 0 \) gives \( y = 4 \), leading us to the point \( (3, 4) \). Since this point is outside our region \( S \) (a disk of radius 1), there are no interior critical points to consider.

Polar Coordinates

Polar coordinates simplify the examination of points along circular boundaries. For the boundary defined by \( x^2 + y^2 = 1 \), expressing these in polar coordinates is efficient:

• Let \( x = \cos\theta \) and \( y = \sin\theta \).

Using these expressions translates the original function into an angle-dependent form. The transformation yields:\[f(\theta) = \cos^2\theta - 6\cos\theta + \sin^2\theta - 8\sin\theta + 7\]Using the identity \( \cos^2\theta + \sin^2\theta = 1 \), the function simplifies to:\[f(\theta) = 1 - 6\cos\theta - 8\sin\theta + 7\]This simplifies the task of evaluating the function anywhere along the circular boundary.

Critical Points

Critical points on a function occur where the derivative is zero or undefined, signifying potential maxima, minima, or saddle points. After converting the boundary function into polar form, find the derivative:\[f'(\theta) = 6\sin\theta - 8\cos\theta\]

• Setting \( f'(\theta) = 0 \) gives insights into points where the rate of change in \( f \) concerning \( \theta \) is zero.

To solve \( 6\sin\theta - 8\cos\theta = 0 \), rearrange to find \( \tan\theta = \frac{8}{6} \). Solving for \( \theta \) provides critical angles:

• At these angles, evaluate the function \( f(\theta) \) to pinpoint specific values of interest (potential max/min).

Global Maximum and Minimum

Finding the global maximum and minimum entails evaluating function values at critical points and key angles. After computing critical angles with \( \tan\theta = \frac{4}{3} \), cross-check with strategic angles:

• Solve at \( \theta = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2} \) for comparison.

For these angles, calculate:

• \( x = \cos\theta \) and \( y = \sin\theta \) pairs.
• Substitute back into the original function \( f(x, y) \) to find values.

Compare values to identify the highest (global maximum) and lowest (global minimum) on the boundary.The comprehensive analysis ensures no critical point is overlooked, establishing reliable extrema values. This process of evaluation can be vital in various applications, from engineering to economics, where understanding absolute optima is paramount.