   Chapter 14.7, Problem 30E

Chapter
Section
Textbook Problem

Use a graphing device as in Example 4 (or Newton’s method or solve numerically using a calculator or computer) to find the critical points of f correct to three decimal places. Then classify the critical points and find the highest or lowest points on the graph, if any.30. f ( x , y )   = 20 e − x 2   −   y 2 sin 3 x cos 3 y , |x| ≤ 1, |y| ≤ 1

To determine

To find: The critical points of the function f(x,y)=20ex2y2sin3xcos3y with |x|1,|y|1 and mention highest or lowest point.

Explanation

Result used:

Second Derivative Test:

“Suppose the second partial derivatives of f are continuous on a disk with center (a,b) , and suppose that fx(a,b)=0 and fy(a,b)=0 (that is (a,b) is a critical point of f).

Let D=D(a,b)=fxx(a,b)fyy(a,b)[fxy(a,b)]2

(a) If D>0 and fxx(a,b)>0 , then f(a,b) is a local minimum.

(b) If D>0 and fxx(a,b)<0 , then f(a,b) is a local maximum.

(c) If D<0 , then f(a,b) is not a local maximum or minimum and it is called a saddle point”.

Given:

The function is, f(x,y)=20ex2y2sin3xcos3y .

Calculation:

Take the partial derivative in the given function with respect to x and obtain fx .

fx=x[20ex2y2sin3xcos3y]=20(ex2y2)(2x)(sin3xcos3y)+20ex2y2(cos3xcos3y)(3)=20ex2y2cos3y(3cos3x2xsin3x)

Thus, fx=20ex2y2cos3y(3cos3x2xsin3x) . (1)

Take the partial derivative in the given function with respect to y and obtain fy .

fy=y[20ex2y2sin3xcos3y]=20(ex2y2)(2y)(sin3xcos3y)+20ex2y2(sin3x(sin3y))(3)=20ex2y2sin3x(3sin3y+2ycos3y)

Thus, fy=20ex2y2sin3x(3sin3y+2ycos3y) . (2)

Solve the equations (1) and (2) and find the values of x and y.

From the equation (1),

20ex2y2cos3y(3cos3x2xsin3x)=0cos3y=0or(3cos3x2xsin3x)=0

From the result, there are two possibilities such either cos3y=0or(3cos3x2xsin3x)=0 .

Suppose, cos3y=0 then,

3y=cos1(0)3y=±π2(|y|1)y=±π6y=±0.524

Suppose (3cos3x2xsin3x)=0 then,

Find the value of x by using the graph.

Use online graphing calculator and draw the graph of the equation (3cos3x2xsin3x)=0 as shown below in Figure 1.

From Figure 1 it can be observed that the values of x are 0.430and0.430 .

From equation (2),

20ex2y2sin3x(3sin3y+2ycos3y)=0sin3x=0or(3sin3y+2ycos3y)=0

From the result, there are two possibilities such either sin3x=0or(3sin3y+2ycos3y)=0

Suppose sin3x=0 then,

sin3x=03x=sin1(0)3x=0x=0

Consider (3sin3y+2ycos3y)=0 .

Find the value of y by using the graph.

Use the online graphing calculator and draw the graph of the equation (3sin3y+2ycos3y)=0 as shown below in Figure 2.

From Figure 1 it can be observed that the values of y are 0,0.872and0.872 .

Therefore, the critical points of the function f(x,y)=20ex2y2sin3xcos3y are (±0.430,0),(0.430,±0.872),(0.430,±0.872)and(0,±0.524) .

Obtain the second derivative, fxx as follows.

Take the partial derivative of the equation (1) with respect to x and obtain fxx .

2fx2=x[20ex2y2cos3y(3cos3x2xsin3x)]=(20cos3y)x[ex2y2(3cos3x)ex2y2(2xsin3x)]=(20cos3y)[ex2y2(3(3)sin3x)+ex2y2(2x)(3cos3x)ex2y2(2x)(2xsin3x)ex2y2(2x(3)cos3x)ex2y2(2sin3x)]=20ex2y2cos3y[(4x211)sin3x12xcos3x]

Hence, 2fx2=20ex2y2cos3y[(4x211)sin3x12xcos3x] .

Take the partial derivative of the equation (2) with respect to y and obtain fyy .

2fy2=y[20ex2y2sin3x(3sin3y+2ycos3y)]=(20sin3x)y[ex2y23sin3y+ex2y22ycos3y]=(20sin3x)[ex2y2(2y)(3sin3y)+ex2y2(3(3)sin3y)+ex2y2(2y)(2ycos3y)+ex2y2(2cos3y)+ex2y2(2y(3)sin3y)]=(20ex2y2sin3x)[(4y211)cos3y12ysin3y]

Hence, 2fy2=(20ex2y2sin3x)[(4y211)cos3y12ysin3y] .

Take the partial derivative of the equation (1) with respect to y and obtain fxy .

2fxy=y[20ex2y2cos3y(3cos3x2xsin3x)]=(3cos3x2xsin3x)y[20ex2y2cos3y]=(3cos3x2xsin3x)[20ex2y2(3sin3y)+20ex2y2(2y)(cos3y)]=20ex2y2(3cos3x2xsin3x)(3sin3y+2ycos3y)

Hence, 2fxy=20ex2y2(3cos3x2xsin3x)(3sin3y+2ycos3y) .

Use second derivative test and obtain the value of D for the critical point (0.430,0) .

fxx(0.430,0)fyy(0.430,0)[fxy(0.430,0)]2}    ={20ex2y2cos3y[(4x211)sin3x12xcos3x](20ex2y2sin3x)[(4y211)cos3y12ysin3y][20ex2y2(3cos3x2xsin3x)(3sin3y+2ycos3y)]2}    ={20e(0.43)2(0)2cos3(0)[(4(0.43)211)sin3(0.43)12(0.43)cos3(0.43)](20e(0.43)2(0)2sin3(0.43))[(4(0)211)cos3(0)12(0)sin(0)][20e(0.43)2(0)2(3cos3(0.43)2(0.43)sin3(0.43))(3sin3(0)+2(0)cos3(0))]2}    ={(16.623)(1)[(10.2604)(sin3(0.43))+12(0.43)cos3(0.43)](20e(0.43)2(0)2(sin3(0.43)))[(4(0)211)cos3(0)12(0)sin(0)][20e(0.43)2(0)2(3cos3(0.43)2(0.43)(sin3(0.43)))(3sin3(0)+2(0)cos3(0))]2}    =368.7187

Since D(±0.430,0)>0 and fxx(0.430,0)>0 there exist the local minimum at (0.430,0) .

Substitute (x,y)=(0.430,0) in f(x,y)=20ex2y2sin3xcos3y ,

f(0.430,0)=20e(0.430)2(0)2[sin3(0.430)][cos3(0)]=20e0.1849[0.0225](1)=20(0.83119)(0.0225)=15.973

Similarly, obtain the value of D for the critical point (0.430,0) .

fxx(0.430,0)fyy(0.430,0)[fxy(0.430,0)]2}={20ex2y2cos3y[(4x211)sin3x12xcos3x](20ex2y2sin3x)[(4y211)cos3y12ysin3y][20ex2y2(3cos3x2xsin3x)(3sin3y+2ycos3y)]2}={20e(0.43)2(0)2cos3(0)[(4(0.43)211)sin3(0.43)12(0.43)cos3(0.43)](20e(0.43)2(0)2sin3(0.43))[(4(0)211)cos3(0)12(0)sin(0)][20e(0.43)2(0)2(3cos3(0.43)2(0.43)sin3(0.43))(3sin3(0)+2(0)cos3(0))]2}={(16.623)(1)[(10.2604)sin3(0.43)12(0.43)cos3(0.43)](20e(0.43)2(0)2sin3(0.43))[(4(0)211)cos3(0)12(0)sin(0)][20e(0.43)2(0)2(3cos3(0.43)2(0.43)sin3(0.43))(3sin3(0)+2(0)cos3(0))]2}=368.7187

Since, D(0.430,0)>0 and fxx<0 there exist a local maximum at (0.430,0) .

Substitute (x,y)=(0.430,0) in f(x,y)=20ex2y2sin3xcos3y ,

f(0.430,0)=20e(0.430)2(0)2[sin3(0.430)][cos3(0)]=20e0.1849[0.0225](1)=20(0.83119)(0.0225)15.973

Obtain the value of D for the critical point (0.430,0.872) .

fxx(0.430,0.872)fyy(0.430,0.872)[fxy(0.430,0.872)]2}    ={20ex2y2cos3y[(4x211)sin3x12xcos3x](20ex2y2sin3x)[(4y211)cos3y12ysin3y][20ex2y2(3cos3x2xsin3x)(3sin3y+2ycos3y)]2}    ={20e(0.43)2(0.872)2cos3(0.872)[(4(0.43)211)sin3(0.43)12(0.43)cos3(0.43)](20e(0.43)2(0.872)2sin3(0.43))[(4(0.872)211)cos3(0.872)12(0.872)sin(0.872)][20e(0.43)2(0.872)2(3cos3(0.43)2(0.43)sin3(0.43))(3sin3(0.872)+2(0.872)cos3(0.872))]2}    ={(16.55746)(2.80556)(16.327)2}    =220.1233

Since, D(0.430,0.872)<0 and fxx(0.430,0.872)>0 there exist the local minimum at (0.430,0.872) .

Substitute (x,y)=(0.430,0.872) in f(x,y)=20ex2y2sin3xcos3y ,

f(0.430,0.872)=20e(0.430)2(0.872)2[sin3(0.430)][cos3(0.872)]=20e0.1849[0.0225](1)=20(0

Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

In Exercises 4756, solve the given equation for the indicated variable. 81=3x

Finite Mathematics and Applied Calculus (MindTap Course List)

In Exercises 6372, evaluate the expression. 63. |6 + 2|

Applied Calculus for the Managerial, Life, and Social Sciences: A Brief Approach

In problems 1-16, solve each equation. 10. Solve

Mathematical Applications for the Management, Life, and Social Sciences

Reduce to lowest terms. 9.

Contemporary Mathematics for Business & Consumers

ex(ex)2 = a) ex3 b) e3x c) 3ex d) e4x

Study Guide for Stewart's Single Variable Calculus: Early Transcendentals, 8th 