Problem 13
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 \(D .\) $$ f(x, y)=2 x+y $$
Step-by-Step Solution
Verified Answer
Global maximum is 3 at (1, 1) and global minimum is -3 at (-1, -1).
1Step 1: Understanding the Domain
The domain \( D \) is a square which includes the points from \((-1, -1)\) to \((1, 1)\) in the plane. Every corner, edge, and interior points within this square are where the function \( f(x, y) = 2x + y \) needs to be evaluated for potential maxima and minima.
2Step 2: Evaluate Function at Corners of D
The function \( f(x, y) = 2x + y \) can have extreme values at the corners of \( D \). We calculate \( f(x, y) \) at each corner: 1. \( f(-1, -1) = 2(-1) + (-1) = -3 \) 2. \( f(-1, 1) = 2(-1) + 1 = -1 \) 3. \( f(1, -1) = 2(1) + (-1) = 1 \) 4. \( f(1, 1) = 2(1) + 1 = 3 \)
3Step 3: Evaluate Function on Edges of D
Next, evaluate \( f(x, y) = 2x + y \) along the edges of \( D \). - For edge \( y = -1 \), \( (-1 \leq x \leq 1) \), \( f(x, -1) = 2x - 1 \). - For edge \( y = 1 \), \( (-1 \leq x \leq 1) \), \( f(x, 1) = 2x + 1 \).- For edge \( x = -1 \), \( (-1 \leq y \leq 1) \), \( f(-1, y) = -2 + y \).- For edge \( x = 1 \), \( (-1 \leq y \leq 1) \), \( f(1, y) = 2 + y \).The maximum and minimum are found at corner values already calculated in Step 2.
4Step 4: Evaluate Function Inside D
The function \( f(x, y) = 2x + y \) is linear, so there are no critical points inside the domain \( D \) other than along the boundaries where it was already evaluated. The extremum values for a linear function on a closed and bounded set occur at boundary points, including corners.
5Step 5: Confirm Global Maxima and Minima
From the evaluations, the maximum value of \( f(x, y) \) is 3 at \((1, 1)\) and the minimum value is -3 at \((-1, -1)\). No greater or lesser values exist inside or on the boundary of \( D \).
Key Concepts
Rectangular DomainEvaluation of FunctionLinear FunctionExtreme Values
Rectangular Domain
When we talk about a rectangular domain, we're referring to a specific area on the coordinate plane. Here, the domain is defined as \( D = \{ (x, y) : -1 \leq x \leq 1, -1 \leq y \leq 1 \} \). This means we're considering a square section of the plane with corners at \((-1, -1)\), \((-1, 1)\), \((1, -1)\), and \((1, 1)\).
To find global maxima and minima of a function over this domain, we need to consider all points including these corners, the edges connecting them, and even the points inside the square.
Each of these parts might offer potential for extreme values, making understanding the full scope of the domain crucial for evaluating the function effectively.
To find global maxima and minima of a function over this domain, we need to consider all points including these corners, the edges connecting them, and even the points inside the square.
Each of these parts might offer potential for extreme values, making understanding the full scope of the domain crucial for evaluating the function effectively.
Evaluation of Function
Evaluating the function involves calculating its values at specific points. For the linear function given as \(f(x, y) = 2x + y\), start by looking at the corners of the domain.
These points are the easiest to compute and can often represent extreme values. In our example, calculating \(f\) at the corners gives:
This step is crucial to affirm that no extreme values are missed that might lie on the boundaries of the domain.
These points are the easiest to compute and can often represent extreme values. In our example, calculating \(f\) at the corners gives:
- \((-1, -1)\) resulting in \(f(-1, -1) = -3\)
- \((-1, 1)\) leading to \(f(-1, 1) = -1\)
- \((1, -1)\) resulting in \(f(1, -1) = 1\)
- \((1, 1)\) leading to \(f(1, 1) = 3\)
This step is crucial to affirm that no extreme values are missed that might lie on the boundaries of the domain.
Linear Function
A linear function, such as \(f(x, y) = 2x + y\), represents a plane when graphed in three dimensions.
These types of functions are the simplest form of multivariable functions, making evaluations straightforward.
Since linear functions do not curve, any extreme values they possess in a bounded domain must occur at the boundary.
This property significantly simplifies the process of finding global maxima and minima, as one does not need to search the interior of the domain \(D\) for critical points like you would with curved functions (e.g., quadratic).
Understanding this property saves significant time in problem-solving, especially in cases involving large domains.
These types of functions are the simplest form of multivariable functions, making evaluations straightforward.
Since linear functions do not curve, any extreme values they possess in a bounded domain must occur at the boundary.
This property significantly simplifies the process of finding global maxima and minima, as one does not need to search the interior of the domain \(D\) for critical points like you would with curved functions (e.g., quadratic).
Understanding this property saves significant time in problem-solving, especially in cases involving large domains.
Extreme Values
In mathematical terms, extreme values refer to the highest and lowest values a function takes within a given domain.
For the function \(f(x, y) = 2x + y\) on the domain \(D\), the extreme values are determined by evaluating the function at the boundaries.
In this case, these values are:
Understanding the extremum concepts is vital in optimization problems, where finding the maximum and minimum values efficiently can lead to solving larger, complex mathematical models.
For the function \(f(x, y) = 2x + y\) on the domain \(D\), the extreme values are determined by evaluating the function at the boundaries.
In this case, these values are:
- Global maximum: \(f(1, 1) = 3\)
- Global minimum: \(f(-1, -1) = -3\)
Understanding the extremum concepts is vital in optimization problems, where finding the maximum and minimum values efficiently can lead to solving larger, complex mathematical models.
Other exercises in this chapter
Problem 13
Evaluate each function at the given point. \(h(x, t)=\exp \left[-\frac{(x-2)^{2}}{2 t}\right]\) at \((1,5)\)
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Show that \(f(x, y)\) is differentiable at the indicated point. \(f(x, y)=\cos (x+y) ;(0,0)\)
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Find \(\frac{d y}{d x}\) if \(y=\arccos x\).
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In the Nicholson-Bailey model, the fraction of hosts escaping parasitism is given by $$ f(P)=e^{-a P} $$ (a) Graph \(f(P)\) as a function of \(P\) for \(a=0.1\)
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