Problem 14 - 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 \(D\). \(f(x, y)=3-x+2 y\)

Step-by-Step Solution

Answer

The absolute maximum is 6, and the absolute minimum is 0.

Step 1: Determine Critical Points Inside D

The function given is linear, meaning it doesn't have any curved features - no hills or valleys. Therefore, we can say that a linear function has no critical points inside the domain because its slope doesn't change unless the partial derivatives are zero everywhere. Calculate the partial derivatives: \(f_x = -1\) and \(f_y = 2\), both of which are constants. This confirms there are no critical points inside \(D\).

Step 2: Evaluate Function at the Corners of D

Since the function is linear, the potential for maxima and minima will happen at the boundaries or corners of the domain. The corners of the given domain \(D\) are the points \((-1, -1)\), \( (-1, 1)\), \((1, -1)\), and \((1, 1)\). Evaluate the function at each of these corners:\[\begin{align*}f(-1,-1) &= 3 - (-1) + 2(-1) = 2,\f(-1,1) &= 3 - (-1) + 2(1) = 6, \f(1,-1) &= 3 - 1 + 2(-1) = 0, \f(1,1) &= 3 - 1 + 2(1) = 4.\end{align*}\]

Step 3: Determine Absolute Maxima and Minima

From our calculations in Step 2, we observe the following values at the corners: \(2, 6, 0,\) and \(4\). The largest value among these is \(6\), and the smallest value is \(0\). Thus, the absolute maximum value of \(f\) occurs at \((-1, 1)\), and the absolute minimum value of \(f\) occurs at \((1, -1)\).

Key Concepts

Understanding Linear Function

A linear function is essentially the simplest type of mathematical function. It is defined by the equation of a line, usually in the form of \( y = mx + b \) for two variables \( x \) and \( y \). In multi-variable calculus, a similar concept applies. The given function \( f(x, y) = 3 - x + 2y \) is linear because each term is either a constant or a constant multiplied by \( x \) or \( y \). This simplicity has a significant implication: the function graph is a flat plane, with no curves or inflections.

Linear functions mean:

• There are no peaks or valleys on their graphs.
• The rate of change, or slope, is constant everywhere.
• Computing derivatives will yield constants, as seen in the solution.

Understanding linear functions helps us know why, in this exercise, critical points do not exist within the domain other than at the boundaries.

Exploring Partial Derivatives

Partial derivatives are like regular derivatives, but for functions of two or more variables like \( f(x, y) \). They measure the rate at which the function changes with respect to one variable, while the other variable is held constant. If you're analyzing how \( f \) changes with respect to \( x \), you're looking at the partial derivative \( f_x \). Similarly, \( f_y \) shows change with respect to \( y \).

In our linear function \( f(x, y) = 3 - x + 2y \), calculating the partial derivatives gives us constants:

• \( f_x = -1 \)
• \( f_y = 2 \)

These values imply the following:

• The function decreases at a constant rate of \(-1\) as \( x \) increases.
• The function increases at a constant rate of \(2\) as \( y \) increases.

The constant nature of these derivatives reaffirms that the function's slope does not change, leading to no critical points within the domain.

Understanding a Rectangular Domain

The domain of a function refers to all possible input values (combinations of \( x \) and \( y \)) where the function is defined. A "rectangular domain" suggests that these combinations are bounded by a rectangle in the plane.

In the original exercise, the rectangular domain is defined as:
\[ D = \{ (x, y) : -1 \leq x \leq 1, -1 \leq y \leq 1 \} \]
This means:

• \( x \) varies from \(-1\) to \(1\).
• \( y \) does the same: \(-1\) to \(1\).

A rectangular domain includes the interior and edges of a rectangle, meaning that for exploring maxima and minima, one needs to consider the boundary and corner points. As this exercise notes, due to the nature of a linear function, the extrema must lie on these boundaries or at the corners. The real charm of studying functions over rectangular domains is in understanding the role of boundaries in optimization problems.