Problem 21 - Calculus for Biology and Medicine

Question

Find the global maxima and minima of $$f(x, y)=x^{2}+y^{2}+4 x-1$$ on the disk $$ D=\left\\{(x, y): x^{2}+y^{2} \leq 9\right\\} $$

Step-by-Step Solution

Answer

The global maxima is 20 at (3, 0); the global minima is -5 at (-2, 0).

Step 1: Identify Critical Points in the Interior

To find critical points of the function on the disk, first compute the partial derivatives: \( f_x = 2x + 4 \) and \( f_y = 2y \). Set these partial derivatives to zero to find critical points. From \( 2x + 4 = 0 \), we get \( x = -2 \). From \( 2y = 0 \), we get \( y = 0 \). Thus, the critical point is \((-2, 0)\).

Step 2: Evaluate the Function at the Critical Points

Calculate the value of the function at the critical point \((-2, 0)\):\[ f(-2, 0) = (-2)^2 + 0^2 + 4(-2) - 1 = 4 - 8 - 1 = -5. \]

Step 3: Consider the Boundary of the Disk

To find extrema on the boundary \(x^2 + y^2 = 9\), parameterize the boundary as \((x, y) = (3\cos\theta, 3\sin\theta)\) where \(0 \leq \theta < 2\pi\). Substitute this into the function: \[ f(3\cos\theta, 3\sin\theta) = (3\cos\theta)^2 + (3\sin\theta)^2 + 4(3\cos\theta) - 1. \]Simplify using \( \cos^2\theta + \sin^2\theta = 1\): \[ f(3\cos\theta, 3\sin\theta) = 9 + 12\cos\theta - 1. \]Thus, \[ g(\theta) = 8 + 12\cos\theta. \]

Step 4: Find the Extrema of the Function on the Boundary

To find the extrema of \(g(\theta)\), note that the cosine function achieves extrema at \(\theta = 0\) and \(\theta = \pi\), corresponding to the maximum and minimum values, respectively. Calculate:- Maximum: \(g(0) = 8 + 12 = 20\).- Minimum: \(g(\pi) = 8 - 12 = -4\).

Step 5: Compare Values to Determine Global Extrema

Compare the function's value at the critical point and its values on the boundary:- Interior: \(f(-2, 0) = -5\).- Boundary Maximum: \(f(3, 0) = 20\).- Boundary Minimum: \(f(-3, 0) = -4\).The global maxima is 20 at points \((3, 0)\), and the global minima is -5 at \((-2, 0)\).

Key Concepts

Critical Points

Critical points are essential in finding the extreme values of a multivariable function. In our exercise, critical points are determined by taking partial derivatives of the function with respect to each variable. The critical points occur where these derivatives are zero.

When we have a function like \[f(x, y)=x^{2}+y^{2}+4x-1,\] we take the partial derivative with respect to \(x\) and \(y\). For instance,

• The partial derivative with respect to \(x\) gives us: \[f_x = 2x + 4.\]
• The partial derivative with respect to \(y\) gives us: \[f_y = 2y.\]

Setting these derivatives to zero helps find critical points:

• Solve \(2x + 4 = 0\) to get \(x = -2\).
• Solve \(2y = 0\) to get \(y = 0\).

Thus, the critical point is \((-2, 0)\). It’s where the function has a potential maximum, minimum, or saddle point, depending on further analysis.

Partial Derivatives

Partial derivatives represent the rate of change of a function concerning one variable while keeping others constant. They are fundamental tools in multivariable calculus because they help analyze how the function behaves in different directions.

In this exercise, we computed the partial derivatives of \[f(x, y)=x^{2}+y^{2}+4x-1.\]

• For **\(x\)**, this meant treating \(y\) as a constant, leading to: \[f_x = 2x + 4.\]
• For **\(y\)**, this involved keeping \(x\) constant, resulting in: \[f_y = 2y.\]

By setting each of these partial derivatives to zero, we find the critical points where the function may change direction.
Partial derivatives are step one in identifying where the function's maximum or minimum values may lie.

Parameterization

Parameterization is a technique used to express a geometric object using equations of one or more parameters. It's especially useful when dealing with curves or boundaries, like the disk in this exercise.

The boundary of our disk was expressed using parameterization in terms of polar coordinates: \((x, y) = (3\cos\theta, 3\sin\theta)\).

• This transformation turns the boundary condition \(x^2 + y^2 = 9\)into its polar form where \(r = 3\).
• This parameterization simplifies calculating the boundary values because angles like \(\theta = 0\)or \(\theta = \pi\)correspond to straightforward cosine and sine values.

Hence, parameterization allows us to reduce complex surfaces or boundaries to simpler expressions in terms of a single variable, here \(\theta\), which represents an angle from the center of the disk to the boundary. It's an approach that turns multivariable challenges into more manageable ones.