Problem 23 - Calculus for Biology and Medicine

Question

Find the absolute 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

Absolute minimum is -1/2 at (-1/2, 1/2) and absolute maximum is 0 on the boundary.

Step 1: Understanding the Problem

We need to find the absolute maxima and minima of \( f(x, y) = x^2 + y^2 + x - y \) on the disk \( x^2 + y^2 \leq 1 \). This involves finding the critical points inside the disk and on its boundary.

Step 2: Finding Critical Points Inside the Disk

To find critical points, we calculate the partial derivatives and set them to zero. The partial derivative with respect to \( x \) is \( f_x = 2x + 1 \), and with respect to \( y \) is \( f_y = 2y - 1 \). Solving \( f_x = 0 \) gives \( x = -\frac{1}{2} \) and \( f_y = 0 \) gives \( y = \frac{1}{2} \). The critical point is \( \left(-\frac{1}{2}, \frac{1}{2}\right) \). This point lies inside the disk since \( \left(-\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 = \frac{1}{2} \leq 1 \).

Step 3: Checking the Function Value at the Critical Point

Substitute \( x = -\frac{1}{2} \) and \( y = \frac{1}{2} \) into the function: \( f\left(-\frac{1}{2}, \frac{1}{2}\right) = \left(-\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right) - \frac{1}{2} = \frac{1}{4} + \frac{1}{4} - \frac{1}{2} - \frac{1}{2} = -\frac{1}{2} \).

Step 4: Boundary Condition Using Lagrange Multipliers

We use Lagrange multipliers to find extrema on the boundary \( x^2 + y^2 = 1 \). The Lagrangian is \( \mathcal{L}(x,y,\lambda) = x^2 + y^2 + x - y + \lambda(1-x^2-y^2) \). The partial derivatives are: \( \frac{\partial \mathcal{L}}{\partial x} = 2x + 1 - 2\lambda x = 0 \), \( \frac{\partial \mathcal{L}}{\partial y} = 2y - 1 - 2\lambda y = 0 \), and \( \frac{\partial \mathcal{L}}{\partial \lambda} = 1 - x^2 - y^2 = 0 \).

Step 5: Solving Lagrange System

From \( \frac{\partial \mathcal{L}}{\partial x} = 0 \), we get \( x = \frac{1}{2(1-\lambda)} \). From \( \frac{\partial \mathcal{L}}{\partial y} = 0 \), \( y = \frac{1}{2(\lambda - 1)} \). Substitute both into \( x^2 + y^2 = 1 \) to solve for \( \lambda \). This gives two solutions: \( x = 0, y = 1 \) and \( x = -1, y = 0 \).

Step 6: Evaluating the Function on the Boundary

Evaluate \( f(x, y) \) at the boundary points: \( f(0, 1) = 0^2 + 1^2 + 0 - 1 = 0 \), and \( f(-1, 0) = (-1)^2 + 0^2 - 1 - 0 = 0 \).

Step 7: Determining Maximum and Minimum Values

The critical value inside is \( f\left(-\frac{1}{2}, \frac{1}{2}\right) = -\frac{1}{2} \), and the function value on both boundary points \( (0,1) \), \( (-1,0) \) is \( 0 \). Therefore, the absolute minimum is \( -\frac{1}{2} \) at \( \left(-\frac{1}{2}, \frac{1}{2}\right) \) and the absolute maximum is \( 0 \) on the boundary.

Key Concepts

absolute maxima and minima

In calculus, absolute maxima and minima are the largest and smallest values a function can take within a given domain. Imagine you're hiking, and you want to find the highest and lowest points on your trail. In math, this involves evaluating all candidates for these extreme values and determining which ones are the most extreme.

When working with functions of two variables, like in this exercise, you must check both the interior of the region and its boundary for these extreme values. To identify potential candidates for absolute maxima and minima:

• Find critical points inside the region where the function's partial derivatives are zero.
• Evaluate the function at these critical points.
• Use other methods, like Lagrange multipliers, to evaluate the function along the boundary of the region.

After evaluating both the interior and boundary, you can identify which values are truly the highest and lowest within the specified region.

critical points

Critical points play a key role in finding extrema (maximum and minimum values) of a function. A critical point occurs where the gradient (or slope) of the function is zero or undefined. In simpler terms, these are spots on a graph where the function stops increasing or decreasing momentarily.

For a function of several variables, like \( f(x, y) = x^2 + y^2 + x - y \), finding the critical points involves calculating the partial derivatives with respect to each variable. These partial derivatives tell us how the function changes in each direction. To find the critical points:

• Set the partial derivative with respect to \( x \), \( f_x \), to zero.
• Set the partial derivative with respect to \( y \), \( f_y \), to zero.

For our function, this leads to solving the equations \( 2x + 1 = 0 \) and \( 2y - 1 = 0 \), resulting in a critical point at \( \left(-\frac{1}{2}, \frac{1}{2}\right) \). In the context of the disk \( x^2 + y^2 \leq 1 \), this point lies within the boundary, making it necessary to evaluate the function value there to determine if it's an absolute maximum or minimum.

Lagrange multipliers

The Lagrange multipliers method is a powerful technique for finding the extrema of functions subject to constraints, like when you're searching for maximum or minimum values along the boundary of a region. It helps navigate the challenges of boundary conditions without explicitly solving them independently.

In this exercise, the constraint is the equation of a circle, \( x^2 + y^2 = 1 \). The function you're optimizing is \( f(x, y) = x^2 + y^2 + x - y \). The method involves setting up a new function called the Lagrangian:

• Form the Lagrangian: \( \mathcal{L}(x, y, \lambda) = f(x, y) + \lambda(g(x, y) - c) \), where \( g(x, y) \) is your constraint \( x^2 + y^2 \), and \( c = 1 \).
• Find the partial derivatives of the Lagrangian with respect to each variable \( x \), \( y \), and the multiplier \( \lambda \), and set them all equal to zero.

This leads to a system of equations that help you find where on the constraint boundary the function could have its maxima or minima. Solving these for our exercise, we found that critical points occur at \( (0,1) \) and \( (-1,0) \), both providing function values equal to zero. This illustrates how the method can streamline calculations and highlight extreme values on boundaries, crucial for finding absolute maxima and minima.