Problem 22 - Calculus for Biology and Medicine

Question

Find the absolute maxima and minima of $$ f(x, y)=x^{2}+y^{2}-6 y+3 $$ on the disk $$ D=\left\\{(x, y): x^{2}+y^{2} \leq 16\right\\} $$

Step-by-Step Solution

Answer

The absolute minimum of \( f \) on \( D \) is \(-6\) at \((0, 3)\). Check boundary points to identify maximum.

Step 1: Calculate Partial Derivatives

Start by finding the first partial derivatives of the function \( f(x, y) = x^2 + y^2 - 6y + 3 \). The partial derivative with respect to \( x \) is \( f_x = 2x \) and with respect to \( y \) is \( f_y = 2y - 6 \).

Step 2: Solve for Critical Points

Set the partial derivatives equal to zero to find critical points: \( 2x = 0 \) gives \( x = 0 \), and \( 2y - 6 = 0 \) gives \( y = 3 \). Thus, the critical point is \( (0, 3) \).

Step 3: Evaluate the Function at the Critical Point

Substitute the critical point back into the function: \( f(0, 3) = 0^2 + 3^2 - 6 \times 3 + 3 = 9 - 18 + 3 = -6 \).

Step 4: Analyze the Boundary

The boundary of the disk \( D \) is defined by \( x^2 + y^2 = 16 \). Substitute \( y = \sqrt{16-x^2} \) into the function and simplify: \( f(x, y) = x^2 + (16 - x^2) - 6\sqrt{16-x^2} + 3 \).

Step 5: Find Critical Points on the Boundary

To find the critical points on the boundary, take the derivative with respect to \( x \) and set it to zero. Simplifying will involve the chain rule and might require solving numerically or by an approximation technique.

Step 6: Evaluate the Function on Boundary

For a more precise calculation, evaluate the function at a few sample points on the boundary, like (4,0), (-4,0), (0,4), and (0,-4), and compare those values to the critical point value.

Step 7: Compare All Values

After evaluating all the critical and boundary values, compare them to find the maximum and minimum values. For instance, if \( f(0, 3) = -6 \), and all boundary evaluations result in higher values, then the absolute minimum could be \(-6\).

Key Concepts

Critical Points

Critical points are locations on a function where the gradient, which consists of partial derivatives, is zero or undefined. In this exercise, the function given is \( f(x, y) = x^2 + y^2 - 6y + 3 \). To find critical points, you calculate its partial derivatives with respect to each variable, \( x \) and \( y \).

These derivatives tell us how the function changes as we move in respective directions. For our function, the partial derivative with respect to \( x \) is \( f_x = 2x \), and with respect to \( y \), it's \( f_y = 2y - 6 \).

You set these derivatives equal to zero, resulting in the system of equations:

• \( 2x = 0 \)
• \( 2y - 6 = 0 \)

Solving these equations gives us the critical point \( (0, 3) \). This point is where the function might reach a local maximum, minimum, or saddle point.

This step is crucial in optimization, as these critical points are candidates for extrema, the highest or lowest values of the function in a given region.

Partial Derivatives

Partial derivatives are foundational in multivariable calculus, especially when dealing with functions of several variables like \( f(x, y) \). They measure the rate of change of a function with respect to one variable, while holding others constant. In our context, for the function \( f(x, y) = x^2 + y^2 - 6y + 3 \), we compute:

• The derivative with respect to \( x \): \( f_x = \frac{\partial f}{\partial x} = 2x \)
• The derivative with respect to \( y \): \( f_y = \frac{\partial f}{\partial y} = 2y - 6 \)

These derivatives simplify the process of understanding how the function behaves as \( x \) or \( y \) changes individually.
Setting these derivatives to zero helps identify potential critical points, which might correspond to peaks, troughs, or stationary points.

Understanding and calculating partial derivatives is vital in optimization problems, as they pave the way toward finding such critical points.

Boundary Analysis

After identifying critical points within the domain, examining the boundary of the region where the function is defined is essential. In our case, the disk \( D = \{(x, y) : x^2 + y^2 \leq 16\} \) defines this region. The boundary is represented by the equation \( x^2 + y^2 = 16 \).

Boundary analysis requires evaluating the function along this edge. One way to do this is by expressing \( y \) in terms of \( x \) (or vice versa). For instance, setting \( y = \sqrt{16 - x^2} \), you substitute back into the original function to evaluate how it behaves on the boundary.

This transforms our task into solving a one-variable optimization problem, which may involve finding critical points using calculus techniques like setting derivatives to zero. This step also considers special points on the boundary, such as intersections, to ensure no potential maximum or minimum is overlooked.
Accurate boundary analysis ensures that the function's global behavior is fully captured within the defined region.

Optimization

Optimization is the process of finding the maximum or minimum values of a function, often under specific constraints. In this problem, the goal is to find absolute maxima and minima of the function \( f(x, y) = x^2 + y^2 - 6y + 3 \) within the disk \( D \).

The process involves:

• Determining and evaluating critical points within the region, as seen with the point \( (0, 3) \).
• Assessing the function along the boundary of the domain \( x^2 + y^2 = 16 \) to check for other potential extrema.

Evaluating the function at sample points along the boundary ensures that no potential extrema are missed. For instance, substituting points like \( (4,0), (-4,0), (0,4), \) and \( (0,-4) \) into the function and checking their values against the internal critical point helps determine the absolute maxima and minima.

Optimization guides decision-making in numerous fields, ensuring that all constraints are considered to find the best possible solution.