Problem 23 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

Global maximum is 1+\sqrt{2}, minimum is 1-\sqrt{2}.

Step 1: Identify the function and region

We are given the function \[ f(x, y) = x^2 + y^2 + x - y \] and need to find its global maxima and minima on the disk \[ D = \{ (x, y) : x^2 + y^2 \leq 1 \} \].The region \(D\) is a closed disk of radius 1 centered at the origin.

Step 2: Find critical points inside the disk

To find critical points, compute the gradient \( abla f = (f_x, f_y) \).Calculate the partial derivatives:\[ f_x = 2x + 1, \quad f_y = 2y - 1 \]Set \( f_x = 0 \) and \( f_y = 0 \) to find critical points:\[ 2x + 1 = 0 \Rightarrow x = -\frac{1}{2} \]\[ 2y - 1 = 0 \Rightarrow y = \frac{1}{2} \]Thus, the critical point inside the disk is \( (-\frac{1}{2}, \frac{1}{2}) \).Check if it lies inside the disk:\[ x^2 + y^2 = \left( -\frac{1}{2} \right)^2 + \left( \frac{1}{2} \right)^2 = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \leq 1 \].Therefore, this point lies within the region \(D\).

Step 3: Evaluate the function at critical points

Evaluate \( f(x, y) = x^2 + y^2 + x - y \) at the critical point:\[ f\left(-\frac{1}{2}, \frac{1}{2}\right) = \left(-\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 - \frac{1}{2} - \frac{1}{2} \]\[ = \frac{1}{4} + \frac{1}{4} - \frac{1}{2} - \frac{1}{2} = -\frac{1}{2} \].

Step 4: Analyze behavior on the boundary

For the boundary \(x^2 + y^2 = 1\), parameterize as \(x = \cos(\theta)\), \(y = \sin(\theta)\).Substitute into \(f(x, y)\):\[ f(\cos(\theta), \sin(\theta)) = \cos^2(\theta) + \sin^2(\theta) + \cos(\theta) - \sin(\theta) \]\[ = 1 + \cos(\theta) - \sin(\theta) \].Find maximum and minimum of \(1 + \cos(\theta) - \sin(\theta)\).

Step 5: Analyze trigonometric expression

Consider: \(1 + \cos(\theta) - \sin(\theta) = 1 + \sqrt{2}\left(\frac{1}{\sqrt{2}}\cos(\theta) - \frac{1}{\sqrt{2}}\sin(\theta)\right)\).This simplifies to \(1 + \sqrt{2}\sin\left(\theta - \frac{\pi}{4}\right)\) due to angle addition formulas.The range of \( \sin \) is \([-1, 1]\), so\[ 1 - \sqrt{2} \leq 1 + \sqrt{2}\sin\left(\theta - \frac{\pi}{4}\right) \leq 1 + \sqrt{2} \].

Step 6: Compare values found

Evaluate the potential minimum and maximum values:1. \(-\frac{1}{2}\) at critical point \((-\frac{1}{2}, \frac{1}{2})\),2. \(1 - \sqrt{2}\) as minimum on the boundary,3. \(1 + \sqrt{2}\) as maximum on the boundary.Thus, the global minimum is \(1 - \sqrt{2}\) and the global maximum is \(1 + \sqrt{2}\).

Key Concepts

Critical Points

To find the global maxima and minima of a function within a region, it is essential first to locate the critical points. These are the points where the gradient of the function becomes zero. The gradient, denoted by \( abla f \), is a vector of partial derivatives of the function with respect to each variable. In simpler terms, it's like finding the slope of the surface at each point.

For our function \( f(x, y) = x^2 + y^2 + x - y \), we calculate the gradient:

• \( f_x = 2x + 1 \)
• \( f_y = 2y - 1 \)

Setting these derivatives to zero, \( f_x = 0 \) and \( f_y = 0 \), allows us to find points where the function is stationary. Solving these equations gives the critical point: \( x = -\frac{1}{2} \) and \( y = \frac{1}{2} \).

Then, we need to check if this point is within our given region, which is inside the disk \( x^2 + y^2 \leq 1 \). By substituting \(x\) and \(y\) back into the equation of the disk, we confirm that it lies within this boundary.

Gradient

The gradient \( abla f \) provides all necessary information to identify the critical points as well as the directional rates of change of the function. It essentially represents the direction of the steepest ascent of the function from a particular point.

For our function, the components are the partial derivatives \( f_x = 2x + 1 \) and \( f_y = 2y - 1 \). These components tell us how sensitive the function is to changes in \(x\) and \(y\).

• At a critical point, the rate of change is zero, indicating a local maximum, local minimum or a saddle point.
• In practice, if \( abla f = (0, 0) \), the function's graph is flat at that point.

Understanding the behavior at these points allows us to predict where the function's values reach limits relative to its neighbors within the space it's defined.

Trigonometric Parameterization

Trigonometric parameterization is a powerful method to simplify finding the behavior of a function along a curve. For boundaries defined by circles or other known shapes, trigonometric identities can reduce complexity by converting variables into angles.

In this problem, we examine the boundary \( x^2 + y^2 = 1 \), a circle of radius 1, by setting \( x = \cos(\theta) \) and \( y = \sin(\theta) \). This reduces the function's form from a two-variable problem into a one-variable trigonometric function:

• \( f(\cos(\theta), \sin(\theta)) = 1 + \cos(\theta) - \sin(\theta) \)

This transformation allows us to use well-known properties of trigonometric functions, such as their maximum and minimum values, to evaluate the boundary's contributions to the global extrema.

Boundary Analysis

In calculus, analyzing the behavior of a function at the boundary of its defined region is crucial to finding global maxima and minima. This involves examining how the function behaves as it approaches the edges of the region.

Once parameterized, our boundary turns into a one-variable function of \( \theta \), giving \( 1 + \cos(\theta) - \sin(\theta) \). To understand its behavior, we use the identity:

• \( 1 + \sqrt{2}\sin(\theta - \frac{\pi}{4}) \)

Since \( \sin \) ranges from -1 to 1, the values of this expression range from \( 1 - \sqrt{2} \) to \( 1 + \sqrt{2} \), determining the function's global minimum and maximum along the boundary.

By evaluating the critical point and the outer boundary behavior, the global extrema are identified as a minimum of \( 1 - \sqrt{2} \) and a maximum of \( 1 + \sqrt{2} \), showcasing how boundary analysis helps confirm findings from inside the region.