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Chapter 13: Problem 5

Find all the local maxima, local minima, and saddle points of the functions. $$f(x, y)=2 x y-x^{2}-2 y^{2}+3 x+4$$

Short Answer

Expert verified
The function has a local maximum at \((3, \frac{3}{2})\).

Step by step solution

01

Find Partial Derivatives

To find the critical points of the function \(f(x, y) = 2xy - x^2 - 2y^2 + 3x + 4\), we first need to calculate the partial derivatives. The partial derivative with respect to \(x\) is \(f_x = \frac{\partial f}{\partial x} = 2y - 2x + 3\). The partial derivative with respect to \(y\) is \(f_y = \frac{\partial f}{\partial y} = 2x - 4y\).
02

Set Derivatives to Zero

Set the partial derivatives \(f_x\) and \(f_y\) to zero to find the critical points. This gives us two equations: 1. \(2y - 2x + 3 = 0\)2. \(2x - 4y = 0\).
03

Solve the System of Equations

Solve the system of equations from Step 2: From equation (2): \(x = 2y\).Substitute into equation (1): \(2y - 2(2y) + 3 = 0\). Simplifying, we find \(2y - 4y + 3 = 0\) which gives \(-2y + 3 = 0\). So \(y = \frac{3}{2}\). Substituting back, \(x = 2(\frac{3}{2}) = 3\). The critical point is \((x, y) = (3, \frac{3}{2})\).
04

Second Partial Derivatives

Find the second partial derivatives: \(f_{xx} = \frac{\partial^2 f}{\partial x^2} = -2\), \(f_{yy} = \frac{\partial^2 f}{\partial y^2} = -4\), \(f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = 2\).
05

Use the Second Derivative Test

Apply the second derivative test to classify the critical point. The Hessian determinant \(D = f_{xx}f_{yy} - (f_{xy})^2\). Substitute values: \(D = (-2)(-4) - (2)^2 = 8 - 4 = 4\). Since \(D > 0\) and \(f_{xx} < 0\), this indicates a local maximum.

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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 essential tools in multivariable calculus. They allow us to analyze how a function changes as one variable is altered while keeping others constant.
For example, given a function of two variables like our exercise: \(f(x, y) = 2xy - x^2 - 2y^2 + 3x + 4\), calculating partial derivatives helps trace the rate of change in specific directions.
  • The partial derivative of \(f\) with respect to \(x\), denoted as \(f_x\), is obtained by differentiating \(f\) with only \(x\), treating \(y\) as a constant. In this function, \(f_x = 2y - 2x + 3\).
  • Likewise, the partial derivative with respect to \(y\), represented as \(f_y\), involves differentiating \(f\) keeping \(x\) as a constant: \(f_y = 2x - 4y\).
By equating these partial derivatives to zero, we set the stage to find critical points where the function's slope in any direction is zero, indicating potential maxima, minima, or saddle points.
Critical Points
Critical points are key to understanding the behavior of a multivariable function. They occur where all partial derivatives equal zero, suggesting potential changes in the function values.
In the given function, \(f_x = 2y - 2x + 3 = 0\) and \(f_y = 2x - 4y = 0\) lead to a system of equations. Solving these equations reveals the critical points: locations where maxima, minima, or saddle points occur.
  • Solve for \(x\) in terms of \(y\): \(x = 2y\).
  • Substitute \(x = 2y\) into the first equation to determine \(y = \frac{3}{2}\).
  • Substitute \(y = \frac{3}{2}\) back to find \(x = 3\).
Thus, the critical point is \((3, \frac{3}{2})\), which is a central figure in exploring the nature of the local extremes.
Second Derivative Test
The second derivative test is an important method to classify critical points. It involves the use of second partial derivatives to determine the nature of extremum at a critical point.
For a two-dimensional problem, we construct the Hessian determinant \(D\) using:
  • \(f_{xx}\) - second partial derivative with respect to \(x\).
  • \(f_{yy}\) - second partial derivative with respect to \(y\).
  • \(f_{xy}\) - mixed partial derivative.
Calculate the Hessian: \(D = f_{xx}f_{yy} - (f_{xy})^2\). If \(D > 0\) and \(f_{xx} < 0\), it indicates a local maximum. In this problem:
  • \(f_{xx} = -2\), \(f_{yy} = -4\), \(f_{xy} = 2\)
  • \(D = (-2)(-4) - (2)^2 = 8 - 4 = 4\)
Since \(D > 0\) and \(f_{xx} < 0\), the critical point \((3, \frac{3}{2})\) is a local maximum.
Local Maxima and Minima
Local maxima and minima are significant in understanding a function's behavior. They help us determine peak points within a localized region of the function, indicating where the function achieves locally highest or lowest values.
At a local maximum, the function value is higher than at all nearby points. This is confirmed when the second derivative test yields \(D > 0\) and \(f_{xx} < 0\).
Conversely, a local minima occurs when the function value is lower than the surrounding points, indicated by \(D > 0\) and \(f_{xx} > 0\).
In our exercise, since the second derivative test confirmed a local maximum for the critical point \((3, \frac{3}{2})\), we conclude that, around this point, function values reach a peak.
Saddle Points
Saddle points are unique critical points that aren't local maxima or minima. The surface of the function resembles a saddle, rising in one direction and falling in another.
They occur where the Hessian determinant \(D < 0\), indicating changes in curvature across various directions.
Saddle points define where a function may "switch" its general direction, not producing locally extreme values.
For the function in question, with the Hessian \(D = 4 > 0\), no saddle point exists at the identified critical point \((3, \frac{3}{2})\). Saddle points provide crucial information about function landscapes, serving as areas where gradients shift forms sharply.

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Most popular questions from this chapter

In economics, the usefulness or utility of amounts \(x\) and \(y\) of two capital goods \(G_{1}\) and \(G_{2}\) is sometimes measured by a function \(U(x, y) .\) For example, \(G_{1}\) and \(G_{2}\) might be two chemicals a pharmaceutical company needs to have on hand and \(U(x, y)\) the gain from manufacturing a product whose synthesis requires different amounts of the chemicals depending on the process used. If \(G_{1}\) costs \(a\) dollars per kilogram, \(G_{2}\) costs \(b\) dollars per kilogram, and the total amount allocated for the purchase of \(G_{1}\) and \(G_{2}\) together is \(c\) dollars, then the company's managers want to maximize \(U(x, y)\) given that \(a x+b y=c .\) Thus, they need to solve a typical Lagrange multiplier problem. Suppose that $$U(x, y)=x y+2 x$$ and that the equation \(a x+b y=c\) simplifies to $$2 x+y=30$$ Find the maximum value of \(U\) and the corresponding values of \(x\) and \(y\) subject to this latter constraint.

Can you conclude anything about \(f(a, b)\) if \(f\) and its first and second partial derivatives are continuous throughout a disk centered at the critical point \((a, b)\) and \(f_{x x}(a, b)\) and \(f_{y y}(a, b)\) differ in sign? Give reasons for your answer.

If you cannot make any headway with \(\lim _{(x, y) \rightarrow(0,0)} f(x, y)\) in rectangular coordinates, try changing to polar coordinates. Substitute \(x=r \cos \theta, y=r \sin \theta,\) and investigate the limit of the resulting expression as \(r \rightarrow 0 .\) In other words, try to decide whether there exists a number \(L\) satisfying the following criterion: $$|r|<\delta \Rightarrow|f(r, \theta)-L|<\epsilon$$ If such an \(L\) exists, then $$\lim _{(x, y) \rightarrow(0,0)} f(x, y)=\lim _{r \rightarrow 0} f(r \cos \theta, r \sin \theta)=L$$ For instance, $$\lim _{(x, y) \rightarrow(0,0)} \frac{x^{3}}{x^{2}+y^{2}}=\lim _{r \rightarrow 0} \frac{r^{3} \cos ^{3} \theta}{r^{2}}=\lim _{r \rightarrow 0} r \cos ^{3} \theta=0$$ To verify the last of these equalities, we need to show that Equation (1) is satisfied with \(f(r, \theta)=r \cos ^{3} \theta\) and \(L=0 .\) That is, we need to show that given any \(\epsilon>0,\) there exists a \(\delta>0\) such that for all \(r\) and \(\theta\) $$|r|<\delta \Rightarrow\left|r \cos ^{3} \theta-0\right|<\epsilon$$ since $$\left|r \cos ^{3} \theta\right|=|r|\left|\cos ^{3} \theta\right| \leq|r| \cdot 1=|r|$$ the implication holds for all \(r\) and \(\theta\) if we take \(\delta=\epsilon\) In contrast, $$\frac{x^{2}}{x^{2}+y^{2}}=\frac{r^{2} \cos ^{2} \theta}{r^{2}}=\cos ^{2} \theta$$takes on all values from 0 to 1 regardless of how small \(|r|\) is, so that \(\lim _{(x, y) \rightarrow(0,0)} x^{2} /\left(x^{2}+y^{2}\right)\) does not exist. In each of these instances, the existence or nonexistence of the limit as \(r \rightarrow 0\) is fairly clear. Shifting to polar coordinates does not always help, however, and may even tempt us to false conclusions. For example, the limit may exist along every straight line (or ray) \(\theta=\) constant and yet fail to exist in the broader sense. Example 5 illustrates this point. In polar coordinates, \(f(x, y)=\left(2 x^{2} y\right) /\left(x^{4}+y^{2}\right)\) becomes $$f(r \cos \theta, r \sin \theta)=\frac{r \cos \theta \sin 2 \theta}{r^{2} \cos ^{4} \theta+\sin ^{2} \theta}$$for \(r \neq 0 .\) If we hold \(\theta\) constant and let \(r \rightarrow 0,\) the limit is \(0 .\) On the path \(y=x^{2},\) however, we have \(r \sin \theta=r^{2} \cos ^{2} \theta\) and $$\begin{aligned} f(r \cos \theta, r \sin \theta) &=\frac{r \cos \theta \sin 2 \theta}{r^{2} \cos ^{4} \theta+\left(r \cos ^{2} \theta\right)^{2}} \\ &=\frac{2 r \cos ^{2} \theta \sin \theta}{2 r^{2} \cos ^{4} \theta}=\frac{r \sin \theta}{r^{2} \cos ^{2} \theta}=1 \end{aligned}$$ Find the limit of \(f\) as \((x, y) \rightarrow(0,0)\) or show that the limit does not exist. $$f(x, y)=\frac{x^{3}-x y^{2}}{x^{2}+y^{2}}$$

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