91Ó°ÊÓ

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 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 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).

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.