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Chapter 12: Problem 46

Find the absolute maximum and minimum values of the following functions on the given region \(R\). \(f(x, y)=6-x^{2}-4 y^{2}\) \(R=\\{(x, y):-2 \leq x \leq 2,-1 \leq y \leq 1\\}\)

Short Answer

Expert verified
Answer: The absolute maximum value is 6, which occurs at (0, 0). The absolute minimum value is -6, which occurs at (-2, -1), (-2, 1), (2, -1), and (2, 1).

Step by step solution

01

Find the Partial Derivatives

To find the partial derivatives, we differentiate with respect to x and y: \(\frac{\partial f}{\partial x} = -2x\) \(\frac{\partial f}{\partial y} = -8y\)
02

Find the Critical Points

To find critical points, we set both partial derivatives equal to zero and solve for x and y: \(-2x = 0\) \(-8y = 0\) From this, we get the critical point: \((x, y) = (0, 0)\)
03

Evaluate the Function at the Critical Points and Boundary Points

We first evaluate the function at the critical point: \(f(0, 0) = 6 - 0^2 - 4\cdot0^2 = 6\) Now we need to evaluate the function along the boundary of the region. There are four sides to the region, so we'll consider each separately: Side 1: \(x = -2\) \(f(-2, y) = 6 - (-2)^2 - 4y^2 = -2 - 4y^2\) \(f(-2, -1) = -6\) \(f(-2, 1) = -6\) Side 2: \(x = 2\) \(f(2, y) = 6 - 2^2 - 4y^2 = -2 - 4y^2\) \(f(2, -1) = -6\) \(f(2, 1) = -6\) Side 3: \(y = -1\) \(f(x, -1) = 6 - x^2 - 4(-1)^2 = 2 - x^2\) \(f(-2, -1) = -2\) \(f(2, -1) = -2\) Side 4: \(y = 1\) \(f(x, 1) = 6 - x^2 - 4(1)^2 = 2 - x^2\) \(f(-2, 1) = -2\) \(f(2, 1) = -2\)
04

Compare the Function Values to Determine Absolute Maximum and Minimum

Now we compare the function values at the critical point and the boundary points: \(f(0, 0) = 6\) \(f(-2, -1) = f(-2, 1) = f(2, -1) = f(2, 1) = -6\) \(f(-2, -1) = f(2, -1) = f(-2, 1) = f(2, 1) = -2\) The absolute maximum is 6, which occurs at (0, 0). The absolute minimum is -6, which occurs at (-2, -1), (-2, 1), (2, -1), and (2, 1).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Derivatives
Partial derivatives are a fundamental concept in multivariable calculus. They measure how a function changes as each variable is slightly adjusted, while all other variables are held constant. For a function of two variables, like our exercise's function, we have two partial derivatives: one for each variable.
  • The partial derivative with respect to a variable is found by differentiating the function while treating other variables as constants.
  • In our example, the function is given as \(f(x, y) = 6 - x^2 - 4y^2\). The partial derivative with respect to \(x\), denoted \( \frac{\partial f}{\partial x} \), is calculated to be \(-2x\), and with respect to \(y\), \( \frac{\partial f}{\partial y} \), is \(-8y\).
These derivatives are crucial for identifying how sensitive the function is to changes in \(x\) and \(y\), and they form the basis for finding critical points.
Critical Points
Critical points of a function occur where the partial derivatives are equal to zero or undefined. This means there isn't any change at these points in the respective direction, contributing to potential maxima or minima.
  • To find critical points, we set the partial derivatives to zero. So in this case, \(-2x = 0\) and \(-8y = 0\) give us one solution: \((x, y) = (0, 0)\).
  • This point is where the slope of the tangent plane is zero or undefined, and therefore, could indicate a local extrema.
  • Determining whether these critical points are indeed maxima, minima, or saddle points requires further analysis, often involving the second derivative test or comparing values as done in our exercise.
This approach allows us to identify key points within the region where the function might possess extreme values.
Boundary Evaluation
Boundary evaluation involves analyzing the function at the edges of the region where it is defined. Sometimes the extrema are not found at internal critical points but along these boundaries.
  • The region \(R\) in this context is defined by \(-2 \leq x \leq 2\) and \(-1 \leq y \leq 1\), creating four sides for us to evaluate.
  • For each boundary, you simplify the function by fixing either \(x\) or \(y\) and analyzing how the other variable impacts the function's value along that border.
  • For instance, along the boundary where \(x = -2\), the function becomes \(f(-2, y) = -2 - 4y^2\), and similarly for other sides.
By evaluating these, we compare the function values at these points and at critical points to determine the absolute extrema, as was performed in the solution.
Multivariable Calculus
Multivariable calculus extends the principles of single-variable calculus to functions of multiple variables. It is essential for analyzing surfaces and spaces in higher dimensions.
  • This branch of mathematics focuses on how these functions behave, change, and optimize, depending on several inputs.
  • Key tools, as seen in our exercise, include partial derivatives and critical points to find local extrema, while boundary evaluation ensures global considerations over the given region.
  • Multivariable calculus also introduces ideas like gradient vectors, which generalize slopes to multiple dimensions, and they play a significant role when the function is complex or has more intricate domains.
Understanding these concepts is vital in fields like physics, engineering, and economics, where systems are rarely isolated to a single variable, and such comprehensive function analysis is required.

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