 # 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 ### Calculus: Early Transcendentals

8th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781285741550 ### Calculus: Early Transcendentals

8th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781285741550

#### Solutions

Chapter
Section
Chapter 14.7, Problem 30E
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

Expert Solution
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 of Solution

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

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