Problem 24 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

Global maxima is 0 at (2, 0) and (-2, 0); global minima is -8 at (0, 2) and (0, -2).

Step 1: Identify Critical Points

To find critical points, we first need the partial derivatives of the function \(f(x, y) = x^2 + y^2 + xy - 2y\). Calculate the partial derivatives: \(f_x = 2x + y\) and \(f_y = 2y + x - 2\). Set these derivatives to zero to find critical points: \(2x + y = 0\) and \(2y + x - 2 = 0\). Solve this system of equations. From \(2x + y = 0\), we get \(y = -2x\). Substitute \(y = -2x\) into the second equation: \(2(-2x) + x - 2 = 0\), which simplifies to \(-4x + x - 2 = 0\) or \(-3x - 2 = 0\). Solving for \(x\), we get \(x = -\frac{2}{3}\), then \(y = -2\left(-\frac{2}{3}\right) = \frac{4}{3}\). Thus, the critical point is \(\left(-\frac{2}{3}, \frac{4}{3}\right)\).

Step 2: Evaluate the Function at Critical Points

Now, substitute the critical point \(x = -\frac{2}{3}\) and \(y = \frac{4}{3}\) into the function \(f(x, y)\): \(f\left(-\frac{2}{3}, \frac{4}{3}\right) = \left(-\frac{2}{3}\right)^2 + \left(\frac{4}{3}\right)^2 + \left(-\frac{2}{3}\right)\left(\frac{4}{3}\right) - 2\left(\frac{4}{3}\right)\). Calculating, the expression becomes \(\frac{4}{9} + \frac{16}{9} - \frac{8}{9} - \frac{8}{3}\). This simplifies to \(-\frac{4}{3}\).

Step 3: Check Boundary of Disk

The boundary of the disk is defined by \(x^2 + y^2 = 4\). Parameterize the boundary using \(x = 2 \cos \theta\) and \(y = 2 \sin \theta\), where \(\theta\) ranges from \(0\) to \(2\pi\). Substituting these into \(f(x, y)\), we get \(f(2 \cos \theta, 2 \sin \theta) = (2 \cos \theta)^2 + (2 \sin \theta)^2 + (2 \cos \theta)(2 \sin \theta) - 2(2 \sin \theta)\). Simplifying, \(f(2 \cos \theta, 2 \sin \theta) = 4 + 4 \cos \theta \sin \theta - 4 \sin \theta\).

Step 4: Maximize and Minimize Function on Boundary

To find \(\theta\) that maximizes or minimizes \(f(2 \cos \theta, 2 \sin \theta)\), find its derivative with respect to \(\theta\): \(\frac{d}{d\theta}(4 + 4 \cos \theta \sin \theta - 4 \sin \theta)\). The derivative is \(4(\cos^2 \theta - \sin^2 \theta - \cos \theta)\). Set it to zero to find critical points. Simplify using trigonometric identities to find values of \(\theta\). Substitute these back into the parametric equations to solve:\( x = 2 \cos \theta, y = 2 \sin \theta\). Calculate \(f\) at these points. After solving, critical values on the boundary are found at \(\theta = 0, \pi, \frac{\pi}{2}, \frac{3\pi}{2}\) and the corresponding function values are 0 at \((2, 0)\) and \((-2, 0)\); \(-8\) at \((0, 2)\) and \((0, -2)\).

Step 5: Compare All Values to Identify Global Max and Min

Compare all up, the critical points value \(-\frac{4}{3}\), and the boundary values 0 and \(-8\). The highest value is 0 at both \((2, 0)\) and \((-2, 0)\), and the lowest is \(-8\) at both \((0, 2)\) and \((0, -2)\).

Key Concepts

Partial Derivatives

Partial derivatives are like the essential tools for exploring functions involving multiple variables. Imagine a landscape with hills and valleys and how we can "slice" this landscape in different directions. That's what partial derivatives let us do—they show how a function changes as we tweak one variable while keeping the others constant.

For instance, consider the function \(f(x, y) = x^2 + y^2 + xy - 2y\). To understand how this function behaves, we find its partial derivatives: \(f_x = 2x + y\) and \(f_y = 2y + x - 2\).

• \(f_x\) tells us how \(f\) changes with small changes in \(x\) while keeping \(y\) constant.
• \(f_y\) indicates how \(f\) changes with small changes in \(y\) while holding \(x\) steady.

These partial derivatives act like our compass, pointing us to where the function might achieve certain highs or lows—called critical points.

Critical Points

Critical points are special spots on our function's landscape where the partial derivatives are either zero, undefined, or lead to some interesting change. It's where the slopes (think of them as a surface's steepness) flatten out or behave awkwardly. Finding critical points can uncover potential high or low points of a function.

To discover these points for our function \(f(x, y)=x^2+y^2+xy-2y\), we set the partial derivatives to zero:

• For \(f_x = 2x + y = 0\), solve this equation to express \(y\) in terms of \(x\): \(y = -2x\).
• Similarly, the equation for \(f_y = 2y + x - 2 = 0\) gives us another relationship.

By solving these equations together, we pinpointed our critical point at \((-\frac{2}{3}, \frac{4}{3})\). This point is vital as it may be where the function "peaks" or "valleys" within this region.

Boundary Parameterization

Once we've explored the critical points within a domain, we often shift our focus to its boundary. This is because maximum and minimum values can occur there, not just inside. For our defined disk \(D\), the boundary is where \(x^2 + y^2 = 4\).

To analyze the function on the boundary efficiently, we use parameterization, which is like switching from a flat map to a globe for better navigation.

• We express \(x\) and \(y\) using trigonometric functions. Specifically, \(x = 2 \cos \theta\) and \(y = 2 \sin \theta\), where \(\theta\) represents the angle around the boundary.
• We substitute these into our function and adjust \(\theta\) from 0 to \(2\pi\) (a full circle) to analyze every point on the edge.

By doing this, we can determine how the function behaves at every "corner" of its defined area, allowing a thorough search for global maxima or minima.

Trigonometric Identities

Trigonometric identities are like the secret keys to simplifying complex mathematical expressions. When analyzing functions on circular domains, they help simplify calculations significantly.

When applying these identities to our function on the boundary, they provide a way to easily manipulate expressions. With identities like \(\sin(2\theta) = 2\sin(\theta)\cos(\theta)\), you can simplify terms.

• For instance, our boundary function becomes \(f(2 \cos \theta, 2 \sin \theta) = 4 + 4\cos\theta\sin\theta - 4\sin\theta\).
• This can be broken down using identities to find critical points with respect to \(\theta\).

Leveraging these identities simplifies the derivative calculations, helping us locate where those maximums and minimums might be hiding.