Problem 16 - Calculus for Biology and Medicine

Question

The functions are defined on the rectangular domain $$ D=\\{(x, y):-1 \leq x \leq 1,-1 \leq y \leq 1\\} $$ Find the absolute maxima and minima of \(f\) on \(D\). \(f(x, y)=x^{2}+y^{2}\)

Step-by-Step Solution

Answer

The absolute minimum is 0 at \((0, 0)\), and the absolute maximum is 2 at the corners \((-1,-1), (-1,1), (1,-1), (1,1)\).

Step 1: Understanding the Problem

We need to find the absolute maximum and minimum values of the function \( f(x, y) = x^2 + y^2 \) within the domain \( D = \{(x, y) : -1 \leq x \leq 1, -1 \leq y \leq 1\} \). This means we evaluate the function's value at points inside and on the boundary of this rectangular domain.

Step 2: Determine Critical Points Inside the Domain

The critical points within the domain are found by setting the partial derivatives equal to zero. Compute \( \frac{\partial f}{\partial x} = 2x \) and \( \frac{\partial f}{\partial y} = 2y \). Setting these partial derivatives to zero gives \( x = 0 \) and \( y = 0 \). So, \((0, 0)\) is a critical point.

Step 3: Evaluate Function at Critical Point

Substitute the critical point \((0, 0)\) back into the function: \( f(0, 0) = 0^2 + 0^2 = 0 \). So, at \( (0, 0) \), the function value is 0.

Step 4: Evaluate Function on the Domain Boundary

For the edges of the domain, evaluate \( f(x, y) \) where: - \( x = -1 \) and \( x = 1 \) with \( y \) ranging from -1 to 1. - \( y = -1 \) and \( y = 1 \) with \( x \) ranging from -1 to 1. For example, when \( x = 1 \), \( f(1, y) = 1^2 + y^2 = 1 + y^2 \). Evaluate at end points \( y = -1 \) and \( y = 1 \), yielding \( 2 \). Similarly for \( x = -1 \), and for edges with fixed \( y \), calculate and find they also reach 2 at all rectangular domain corners.

Step 5: Identify Absolute Maxima and Minima

Based on evaluation:- The minimum value inside \( D \) is 0 at \((0, 0)\).- The maximum value on the edges is 2 at corners \((-1, -1)\), \((-1, 1)\), \((1, -1)\), and \((1, 1)\).

Key Concepts

Critical Points

Critical points of a function are locations in the domain where the function might have a local maximum, a local minimum, or a saddle point. To find these points for a multivariable function like \[ f(x, y) = x^2 + y^2 \] on a rectangular domain, we need to compute the partial derivatives with respect to each variable and set them to zero.Here's what that means:

• Compute the partial derivative with respect to \( x \): \( \frac{\partial f}{\partial x} = 2x \).
• Compute the partial derivative with respect to \( y \): \( \frac{\partial f}{\partial y} = 2y \).
• Set these expressions equal to zero to find: \( 2x = 0 \) and \( 2y = 0 \), giving us \( x = 0 \) and \( y = 0 \).

Thus, the only critical point within this domain is at \( (0, 0) \). This tells us we should check this point when determining the function's maximum and minimum values within the domain. Simple, right? Partial derivatives guide us here by identifying potential points of change within the interior of the domain.

Partial Derivatives

Partial derivatives are a fundamental tool in multivariable calculus, used to study the way a function changes in each direction within its domain. For functions of two variables like \( f(x, y) = x^2 + y^2 \), we use partial derivatives to isolate changes in \( x \) and \( y \) independently.In our example:

• The partial derivative with respect to \( x \), \( \frac{\partial f}{\partial x} = 2x \), helps us see how changes in the \( x \) variable affect the function's value while keeping \( y \) constant.
• Similarly, \( \frac{\partial f}{\partial y} = 2y \) indicates how variations in \( y \) impact the function, holding \( x \) steady.

These derivatives help us find where the function has potential local extremes by setting them to zero. This invaluable insight helps us identify and evaluate the critical points required for optimization. By understanding partial derivatives, we're essentially peering into the sensitivity of functions concerning their variables.

Domain Boundary Evaluation

After locating internal critical points, our function's behavior on the domain boundary must be scrutinized. This means examining the points where the function might achieve its most extreme values along the limits of its domain. For the domain \( D = \{(x, y) : -1 \leq x \leq 1, -1 \leq y \leq 1\} \), we evaluate:

• Edges \( x = -1 \) and \( x = 1 \), with \( y \) ranging from -1 to 1.
• Edges \( y = -1 \) and \( y = 1 \), with \( x \) ranging from -1 to 1.

When evaluating at these edges, substitute values directly into \( f(x, y) = x^2 + y^2 \). It reveals that each of these edges reaches the value 2 at specific points like \( (1, 1) \), given the corners share the maximum distances from zero within the constraints.This boundary evaluation is crucial since the function's extremum might not only be within the interior but also along or at the endpoints of the domain's boundary.

Absolute Maxima and Minima

An absolute maximum or minimum is the highest or lowest point the function reaches over its entire domain, respectively. We've previously identified our critical point at \( (0, 0) \), with a function value of 0. This is called the absolute minimum since no lower values exist for this function within its defined domain.In contrast, by evaluating on the domain’s boundary, we found the maximum function value at 2, occurring at the corners \((-1, -1)\), \((-1, 1)\), \((1, -1)\), and \((1, 1)\).Some highlights to understand:

• The highest value, "maximum," is 2 on the boundary, and it doesn’t get better within the domain.
• The lowest value, "minimum," is 0, an occurrence at \( (0, 0) \) as internally calculated.

Recognizing these points is important for correctly identifying where and how a function behaves in its entirety, which is the essence of optimization. Properly evaluating both critical points and boundaries paints the full picture of function extremity.