/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 10 Assume the second derivatives of... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Assume the second derivatives of \(f\) are continuous throughout the xy-plane and \(f_{x}(0,0)=f_{y}(0,0)=0 .\) Use the given information and the Second Derivative Test to determine whether \(f\) has a local minimum, a local maximum, or a saddle point at \((0,0),\) or state that the test is inconclusive. $$f_{x x}(0,0)=-6, f_{y y}(0,0)=-3, \text { and } f_{x y}(0,0)=4$$

Short Answer

Expert verified
Answer: For the given function, the critical point (0,0) is a local maximum.

Step by step solution

01

Compute the discriminant D

The discriminant is given by the formula: $$D=f_{xx}(x_0,y_0)f_{yy}(x_0,y_0)-(f_{xy}(x_0,y_0))^2$$ Plugging in the given partial derivatives at the point \((0,0)\): $$D=(-6)(-3)-(4)^2=18-16=2$$
02

Use the Second Derivative Test

According to the Second Derivative Test, if \(D>0\) and \(f_{xx}(x_0,y_0)>0\), then the critical point is a local minimum; if \(D>0\) and \(f_{xx}(x_0,y_0)<0\), then the critical point is a local maximum; if \(D<0\), then the critical point is a saddle point; and if \(D=0\), the test is inconclusive. In our case: - \(D=2>0\): Discriminant positive - \(f_{xx}(0,0)=-6<0\): The second partial derivative with respect to \(x\) is negative
03

Make the conclusion

Since \(D>0\) and \(f_{xx}(0,0)<0\), the Second Derivative Test tells us that the critical point \((0,0)\) is a local maximum for the function \(f(x,y)\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Local Maximum and Minimum
When analyzing the graph of a function, particularly in two dimensions or more, a point can be at the peak (highest value in the neighborhood), known as the local maximum, or at the bottom of a valley (lowest value in the neighborhood), known as the local minimum. These are points where the function changes direction, and they provide essential information when understanding the function’s behavior.

Mathematically, if the function has a local maximum at a point, the value of the function at this point is greater than or equal to the value at nearby points. Conversely, for a local minimum, the function value is less than or equal to its neighbors. Determining these points is crucial for optimization problems, like finding the best solution that maximizes profit or minimizes cost in real-world scenarios.
Critical Points
In calculus, critical points are the backbone of finding local maxima and minima. They are where the first derivatives of a function with respect to all variables are zero or undefined. These points are essential because they are potential locations for local maxima or minima.

However, not all critical points correspond to a maximum or minimum. To discern the nature of these points, further analysis such as the Second Derivative Test is often employed. At these junctures, other behaviors such as saddle points – where a point is a minimum along one cross-section and a maximum along another – might also be identified.
Partial Derivatives
Dealing with functions of multiple variables requires the use of partial derivatives, which are derivatives with respect to one variable while holding the others constant. The notations like fx and fy signify the rate at which the function changes as the respective variable changes, while the other remains unchanged.

For example, in the context of a function f(x,y), the partial derivative fx explains how the function value changes in the x-direction and fy in the y-direction. The importance of partial derivatives lies in their role in finding critical points, plotting the function's graph, and analyzing the function’s behavior, such as increasing or decreasing trends in multidimensional spaces.
Discriminant of a Function
When using the Second Derivative Test for functions of two variables, the discriminant serves as a key player. It is a function of the partial derivatives, specifically, given by the formula:
\[D=f_{xx}f_{yy}-(f_{xy})^2\].
If the discriminant is positive and the second partial derivative with respect to x is positive, it indicates a local minimum. If that derivative is negative, it suggests a local maximum. However, if the discriminant is negative, neither a maximum nor minimum is assured, and instead, one typically has a saddle point. The discriminant helps identify the nature of the critical points and guide us to understand the 'curvature' of the function around these points.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Use Lagrange multipliers in the following problems. When the constraint curve is unbounded, explain why you have found an absolute maximum or minimum value. Shipping regulations A shipping company requires that the sum of length plus girth of rectangular boxes not exceed 108 in. Find the dimensions of the box with maximum volume that meets this condition. (The girth is the perimeter of the smallest side of the box.)

Use Lagrange multipliers in the following problems. When the constraint curve is unbounded, explain why you have found an absolute maximum or minimum value. Maximum area rectangle in an ellipse Find the dimensions of the rectangle of maximum area with sides parallel to the coordinate axes that can be inscribed in the ellipse \(4 x^{2}+16 y^{2}=16\)

Among all triangles with a perimeter of 9 units, find the dimensions of the triangle with the maximum area. It may be easiest to use Heron's formula, which states that the area of a triangle with side length \(a, b,\) and \(c\) is \(A=\sqrt{s(s-a)(s-b)(s-c)},\) where \(2 s\) is the perimeter of the triangle.

Looking ahead- tangent planes Consider the following surfaces \(f(x, y, z)=0,\) which may be regarded as a level surface of the function \(w=f(x, y, z) .\) A point \(P(a, b, c)\) on the surface is also given. a. Find the (three-dimensional) gradient of \(f\) and evaluate it at \(P\). b. The set of all vectors orthogonal to the gradient with their tails at \(P\) form a plane. Find an equation of that plane (soon to be called the tangent plane). $$f(x, y, z)=x y+x z-y z-1=0 ; P(1,1,1)$$

Use what you learned about surfaces in Sections 13.5 and 13.6 to sketch a graph of the following functions. In each case, identify the surface and state the domain and range of the function. $$H(x, y)=\sqrt{x^{2}+y^{2}}$$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.