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Chapter 10: Problem 22

Find the global maxima and minima of $$f(x, y)=2 x^{2}+y^{2}-6 y+3$$ on the disk $$ D=\left\\{(x, y): x^{2}+y^{2} \leq 16\right\\} $$

Short Answer

Expert verified
Global maximum is 75; global minimum is -6.

Step by step solution

01

Identify the function and the constraint

The function given is \( f(x, y) = 2x^2 + y^2 - 6y + 3 \). The constraint for the disk is \( x^2 + y^2 \leq 16 \), which represents a circle with radius 4 centered at the origin.
02

Find Critical Points Inside the Disk

To find critical points inside the disk, calculate the partial derivatives: \( f_x = 4x \) and \( f_y = 2y - 6 \). Set them to zero: \( 4x = 0 \) and \( 2y - 6 = 0 \), which gives the solutions \( x = 0 \) and \( y = 3 \). Therefore, a critical point is \( (0, 3) \). Check if this point lies inside the disk by verifying that \( 0^2 + 3^2 \leq 16 \), which it does.
03

Determine Function Value at Critical Points

Calculate \( f(0, 3) \): \( f(0, 3) = 2(0)^2 + (3)^2 - 6(3) + 3 = 9 - 18 + 3 = -6 \). So, the value of the function at the critical point is \( -6 \).
04

Check the Boundary of the Disk

For points on the boundary defined by \( x^2 + y^2 = 16 \), use a parameterization such as \( x = 4\cos\theta \) and \( y = 4\sin\theta \). Substitute into the function: \( f(4\cos\theta, 4\sin\theta) = 2(16\cos^2\theta) + 16\sin^2\theta - 24\sin\theta + 3 \). Simplify using \( \cos^2\theta + \sin^2\theta = 1 \): \( f = 32 + 16 - 24\sin\theta + 3 = 51 - 24\sin\theta \).
05

Find Maximum and Minimum on the Boundary

The expression \( 51 - 24\sin\theta \) is maximum when \( \sin\theta = -1 \) and minimum when \( \sin\theta = 1 \). For maximum: \( 51 - 24(-1) = 75 \). For minimum: \( 51 - 24(1) = 27 \).
06

Compare and Conclude Global Maxima and Minima

Compare function values: Inside the disk, \( f(0, 3) = -6 \); on the boundary, the minimum is 27, and the maximum is 75. Thus, the global maximum is 75 (when \( \sin\theta = -1 \)) and the global minimum is -6 (at \( x = 0, y = 3 \)).

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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 fundamental in finding the rates of change of a function with respect to variables in multivariable calculus. For a function like \( f(x, y) = 2x^2 + y^2 - 6y + 3 \), the partial derivative with respect to \( x \), denoted \( f_x \), represents how \( f \) changes as \( x \) changes, while keeping \( y \) constant.
The partial derivative \( f_x = 4x \) tells us that the function grows or shrinks linearly with \( x \). Similarly, \( f_y = 2y - 6 \) shows how the function changes with \( y \), also affecting the slope of \( f \) at a given point.
This focus on changing variables separately allows us to locate points of interest, like critical points, by setting these derivatives to zero, enabling further analyses of the function's behavior.
Critical Points
Critical points occur where the first derivatives of a function are zero or undefined. These points represent potential locations for local minima or maxima. For the exercise, setting \( f_x = 0 \) and \( f_y = 0 \) led to \( 4x = 0 \) and \( 2y - 6 = 0 \).
Solving these equations gives the critical point \((0, 3)\). This point must be verified within the constraint boundary, which it satisfies as it lies inside the disk \( x^2 + y^2 \leq 16 \).
Once a critical point is identified within the context of the problem, it becomes crucial to evaluate the function's value at this point. In this case, \( f(0, 3) \) calculated to \( -6 \), indicating a potential global minimum within the region of interest.
Parameterization
Parameterization involves expressing the coordinates of a geometric figure using a single parameter. For the boundary of a circular disk, parameterization simplifies analysis by reducing the function of two dimensions to one.
In the exercise, we parameterize the circle's boundary with \( x = 4\cos\theta \) and \( y = 4\sin\theta \). This turns the constraint \( x^2 + y^2 = 16 \) into a manageable form: \( 51 - 24\sin\theta \).
This technique aids in examining the function鈥檚 behavior explicitly on the boundary, allowing us to test for extreme values at specific angles corresponding to the maximum and minimum values possible as \( \sin\theta \) varies from \(-1\) to \(1\).
Constraint
A constraint in problem-solving refers to a condition that the solution must meet. Here, the constraint \( x^2 + y^2 \leq 16 \) defines a closed region (a disk) where the function's extrema are to be found.
Recognizing this constraint helps limit the domain of the function analysis. It dictates that not all potential critical points of \( f(x, y) \) are valid, only those within the given circle.
In optimization problems like this, constraints shape the feasible region but also play a role in imposing limits on derivatives, integral calculations, and potential parameter values throughout the problem-solving process, ultimately guiding the determination of global extrema.

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Most popular questions from this chapter

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