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 global maxima and minima of \(f\) on \(D .\) $$ f(x, y)=3-x+2 y $$

Step-by-Step Solution

Answer

The global maximum is 6 at \((-1, 1)\) and the global minimum is 0 at \((1, -1)\).

Step 1: Understand the Problem

We need to find the global maxima and minima of the function \(f(x, y) = 3 - x + 2y\) within the domain \(D = \{(x, y):-1 \leq x \leq 1, -1 \leq y \leq 1\}\), which is a square defined by its corner points: \((-1, -1), (-1, 1), (1, -1), (1, 1)\).

Step 2: Evaluate Function at Corner Points

The corner points of the domain \(D\) are \((-1,-1), (-1, 1), (1, -1), (1, 1)\). Evaluate \(f(x, y)\) at these points: - \(f(-1, -1) = 3 - (-1) + 2(-1) = 2\)- \(f(-1, 1) = 3 - (-1) + 2(1) = 6\)- \(f(1, -1) = 3 - 1 + 2(-1) = 0\)- \(f(1, 1) = 3 - 1 + 2(1) = 4\).

Step 3: Determine Global Maxima and Minima

Compare the values calculated in Step 2: - Maximum value of \(f\) among these points is 6 (at \((-1, 1)\)) and - Minimum value of \(f\) is 0 (at \((1, -1)\)).

Key Concepts

Global Maxima and Minima

When we talk about global maxima and minima, we are interested in finding the largest and smallest values that a function can take over an entire domain. Here, the function is defined as \[ f(x, y) = 3 - x + 2y \].Within the context of multivariable calculus, examining global maxima and minima involves checking the values that the function can reach at certain critical points. However, for simple linear functions like this one defined over a rectangular domain, we focus our attention to the boundary — especially the corners — since these typically contain the extremal values. By evaluating the function at these extreme points, which forms a finite set, we ascertain both the global maximum and minimum values. In this case, the maximum value obtained is 6, while the minimum value is 0. These values signify the peaks and troughs of elevations mapped out by the function over the specified region.

Rectangular Domain

Multivariable calculus often involves analyzing functions over specific domains. These domains define the range of input values that can be used in the functions. Here, the domain is given as a rectangular area:\[ D = \{(x, y): -1 \leq x \leq 1, -1 \leq y \leq 1\} \].This domain is essentially a square on the coordinate plane with boundary corners at the points:

• iu(-1, -1)
• (-1, 1)
• (1, -1)
• (1, 1)

Each point denotes where the edges of this square intersect. Within this confined space, we often assess the behavior of the function to understand how different value constraints impact the function's output. Analyzing how the function behaves at these points, especially at corners, provides critical insights that lead us to determine where the maximal and minimal values lie.

Function Evaluation

Evaluating a function like \[ f(x, y) = 3 - x + 2y \] involves substituting specific values for \(x\) and \(y\). This step is vital, especially when dealing with boundary analyses in multivariable calculus. Evaluating at different points lets us understand how the function behaves across its domain.To comprehensively evaluate:

• Start by substituting the coordinate values of each corner point into the function.
• Calculate the result to determine what the function equals at each of these points.

For instance, a function evaluation at the corner point \((-1, 1)\) yields\[ f(-1, 1) = 3 - (-1) + 2(1) = 6 \].Through similar evaluations at each corner, as done in the step-by-step solution, we can gather enough information to make inferences about where the function's maximum and minimum values fall within the domain. Evaluating each corner point is key to ensuring we've checked each potential extremal value efficiently.