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Chapter 12: Problem 52

Find the absolute maximum and minimum values of the following functions on the given region \(R\). \(f(x, y)=\sqrt{x^{2}+y^{2}} ; R\) is the closed region bounded by the ellipse \(\frac{x^{2}}{4}+y^{2}=1\).

Short Answer

Expert verified
Based on the given function \(f(x, y) = \sqrt{x^2 + y^2}\) and the closed region bounded by the ellipse \(\frac{x^2}{4} + y^2 = 1\), the absolute minimum value of the function is 0, and the absolute maximum value is \(\sqrt{5}\).

Step by step solution

01

1. Parameterize the boundary

The boundary of the region is the ellipse \(\frac{x^2}{4} + y^2 = 1\). We can parameterize it using polar coordinates, let \(x = 2\cos(\theta)\) and \(y = \sin(\theta)\). Then the boundary curve can be described as \((2\cos(\theta), \sin(\theta))\), where \(0 \leq \theta \leq 2\pi\).
02

2. Find the gradient of the function

The gradient of the function is given by \(\nabla f(x, y) = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right)\). We can compute the partial derivatives as follows: $$ \frac{\partial f}{\partial x} = \frac{x}{\sqrt{x^2 + y^2}} \quad \text{and} \quad \frac{\partial f}{\partial y} = \frac{y}{\sqrt{x^2 + y^2}} $$ So, \(\nabla f(x, y) = \left(\frac{x}{\sqrt{x^2 + y^2}}, \frac{y}{\sqrt{x^2 + y^2}}\right)\).
03

3. Determine the critical points

Critical points occur when \(\nabla f(x, y) = (0, 0)\). Since the region is closed and bounded, we only need to consider the interior. Set each component of the gradient to zero and solve for \((x, y)\): $$ \frac{x}{\sqrt{x^2 + y^2}} = 0 \quad \text{and} \quad \frac{y}{\sqrt{x^2 + y^2}} = 0 $$ The only point that satisfies these equations is \((x, y) = (0, 0)\). Thus, \((0, 0)\) is the only critical point.
04

4. Evaluate the function at the critical points and the boundary

Now we need to find the values of the function at the critical point (\(\mathbf{c} = (0, 0)\)) and along the boundary, which we can substitute the parameterization for \((x, y)\): $$ f(\mathbf{c}) = f(0, 0) = 0 \\ f(2\cos(\theta), \sin(\theta)) = \sqrt{(2\cos(\theta))^2 + \sin^2(\theta)} = \sqrt{4\cos^2(\theta) + \sin^2(\theta)} $$
05

5. Compare the resulting values

To find the absolute maximum and minimum values of the function, we compare the values of the function at the critical point and along the boundary. We have: $$ f(0, 0) = 0 \\ f(2\cos(\theta), \sin(\theta)) = \sqrt{4\cos^2(\theta) + \sin^2(\theta)} $$ Notice that the function value on the boundary is always non-negative (since it's a square root) and equal to zero only at the critical point, which means the absolute minimum is \(f(0, 0) = 0\). To find the maximum, observe that since \(0 \leq 4\cos^2\theta + \sin^2\theta \leq 5\) (by plugging the minimum and maximum values of the trigonometric functions), we have: $$ 0 = \sqrt{0} \leq \sqrt{4\cos^2(\theta) + \sin^2(\theta)} \leq \sqrt{5} $$ So, the maximum can only be achieved at the boundary, and the absolute maximum value is \(\sqrt{5}\). Thus, the absolute minimum value is 0, and the absolute maximum value is \(\sqrt{5}\).

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Key Concepts

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

Gradient of a Function
The gradient of a function is a crucial tool in multivariable calculus. It provides a vector containing all the partial derivatives of the function, pointing in the direction of the steepest ascent. For a two-variable function, such as our function \(f(x, y) = \sqrt{x^2 + y^2}\), the gradient is denoted as \(abla f(x, y)\) and calculated by taking the partial derivatives with respect to each variable.

Specifically, for our \(f\), the gradient is \(abla f(x, y) = \left(\frac{x}{\sqrt{x^2 + y^2}}, \frac{y}{\sqrt{x^2 + y^2}}\right)\). This formula provides a way to determine the rate of change of the function in any direction at a point. In the context of finding absolute maximum and minimum values, the gradient is used to identify critical points, where the function may attain extreme values inside its domain.
Critical Points
In the journey to find extreme values of a function, critical points are like signposts that guide mathematicians to potential highs and lows of a function's landscape. These are points where the function's gradient is zero or the function is not differentiable. For our function \(f(x, y)\), critical points arise when both components of its gradient are zero, meaning \(abla f(x, y) = (0, 0)\).

The only point in our function's domain meeting this criterion is \( (x, y) = (0, 0) \), making it our sole critical point. Since we're assessing a closed region, we know that extreme values can occur at critical points or along the boundary of this region, preparing the stage for further analysis of these specific locations.
Parameterization
The technique of parameterization is translating one type of coordinate system into another, providing a new way to describe a curve or surface in space. Understanding parameterization is crucial for tackling problems involving constrained optimization, like maximizing or minimizing a function under certain conditions.

In our case, we parameterize the boundary of the elliptical region \(R\) using the equations \(x = 2\cos(\theta)\) and \(y = \sin(\theta)\). Here, \(\theta\) becomes the parameter that allows us to express the coordinates \(x\) and \(y\) in terms of a single variable, which is very convenient when evaluating the function along the curve of the ellipse.
Polar Coordinates
The concept of polar coordinates is another important tool, offering an alternative to the Cartesian coordinate system for representing points on a plane. Instead of using perpendicular x and y-axes, polar coordinates define a point's location based on its angle \(\theta\) from the positive x-axis and its distance \(r\) from the origin.

Our exercise involves polar coordinates implicitly as we deal with an ellipse, which can be conveniently parameterized using polar-like equations. While traditional polar coordinates use \(r\) and \(\theta\) to denote a circle's radius and angle, for an ellipse we adjust these to \(x = 2r\cos(\theta)\) and \(y = r\sin(\theta)\), where \(r\) equals 1 to stay within the boundary of ellipse \(\frac{x^2}{4} + y^2 = 1\). This polar approach simplifies the process of evaluating our function along the elliptical boundary.

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Most popular questions from this chapter

Determine whether the following statements are true and give an explanation or counterexample. a. The plane passing through the point (1,1,1) with a normal vector \(\mathbf{n}=\langle 1,2,-3\rangle\) is the same as the plane passing through the point (3,0,1) with a normal vector \(\mathbf{n}=\langle-2,-4,6\rangle\) b. The equations \(x+y-z=1\) and \(-x-y+z=1\) describe the same plane. c. Given a plane \(Q\), there is exactly one plane orthogonal to \(Q\). d. Given a line \(\ell\) and a point \(P_{0}\) not on \(\ell\), there is exactly one plane that contains \(\ell\) and passes through \(P_{0}\). e. Given a plane \(R\) and a point \(P_{0}\), there is exactly one plane that is orthogonal to \(R\) and passes through \(P_{0}\) f. Any two distinct lines in \(\mathbb{R}^{3}\) determine a unique plane. g. If plane \(Q\) is orthogonal to plane \(R\) and plane \(R\) is orthogonal to plane \(S\), then plane \(Q\) is orthogonal to plane \(S\).

Economists use utility functions to describe consumers' relative preference for two or more commodities (for example, vanilla vs. chocolate ice cream or leisure time vs. material goods). The Cobb-Douglas family of utility functions has the form \(U(x, y)=x^{a} y^{1-a},\) where \(x\) and \(y\) are the amounts of two commodities and \(0 < a < 1\) is a parameter. Level curves on which the utility function is constant are called indifference curves; the utility is the same for all combinations of \(x\) and \(y\) along an indifference curve (see figure). a. The marginal utilities of the commodities \(x\) and \(y\) are defined to be \(\partial U / \partial x\) and \(\partial U / \partial y,\) respectively. Compute the marginal utilities for the utility function \(U(x, y)=x^{a} y^{1-a}\) b. The marginal rate of substitution (MRS) is the slope of the indifference curve at the point \((x, y) .\) Use the Chain Rule to show that for \(U(x, y)=x^{a} y^{1-a},\) the MRS is \(-\frac{a}{1-a} \frac{y}{x}\) c. Find the MRS for the utility function \(U(x, y)=x^{0.4} y^{0.6}\) at \((x, y)=(8,12)\)

a. Find the linear approximation to the function \(f\) at the given point. b. Use part (a) to estimate the given function value. $$f(x, y)=\sqrt{x^{2}+y^{2}} ;(3,-4) ; \text { estimate } f(3.06,-3.92).$$

Given three distinct noncollinear points \(A, B,\) and \(C\) in the plane, find the point \(P\) in the plane such that the sum of the distances \(|A P|+|B P|+|C P|\) is a minimum. Here is how to proceed with three points, assuming that the triangle formed by the three points has no angle greater than \(2 \pi / 3\left(120^{\circ}\right)\). a. Assume the coordinates of the three given points are \(A\left(x_{1}, y_{1}\right)\) \(\underline{B}\left(x_{2}, y_{2}\right),\) and \(C\left(x_{3}, y_{3}\right) .\) Let \(d_{1}(x, y)\) be the distance between \(A\left(x_{1}, y_{1}\right)\) and a variable point \(P(x, y) .\) Compute the gradient of \(d_{1}\) and show that it is a unit vector pointing along the line between the two points. b. Define \(d_{2}\) and \(d_{3}\) in a similar way and show that \(\nabla d_{2}\) and \(\nabla d_{3}\) are also unit vectors in the direction of the line between the two points. c. The goal is to minimize \(f(x, y)=d_{1}+d_{2}+d_{3} .\) Show that the condition \(f_{x}=f_{y}=0\) implies that \(\nabla d_{1}+\nabla d_{2}+\nabla d_{3}=0\) d. Explain why part (c) implies that the optimal point \(P\) has the property that the three line segments \(A P, B P,\) and \(C P\) all intersect symmetrically in angles of \(2 \pi / 3\) e. What is the optimal solution if one of the angles in the triangle is greater than \(2 \pi / 3\) (just draw a picture)? f. Estimate the Steiner point for the three points (0,0),(0,1) and (2,0).

Show that the Second Derivative Test is inconclusive when applied to the following functions at(0,0) Describe the behavior of the function at the critical point. \(f(x, y)=x^{2} y-3\)
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