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Multivariable Calculus

8th Edition
James Stewart
ISBN: 9781305266643

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BuyFindarrow_forward

Multivariable Calculus

8th Edition
James Stewart
ISBN: 9781305266643
Textbook Problem

Find the local maximum and minimum values and saddle point(s) of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function.

19. f(x, y) = y2 − 2y cos x, −1 ≤ x ≤ 7

To determine

To find: The local maximum, local minimum and saddle point of the function f(x,y)=y22ycosx , 1x7 .

Explanation

Given:

The function is, f(x,y)=y22ycosx .

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”.

Calculation:

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

fx=x(y22ycosx)=y2x(1)2yx(cosx)=0+2ysinx=2ysinx

Thus, fx=2ysinx (1)

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

fy=y(y22ycosx)=y(y2)2cosxy(y)=2y2cosx(1)=2y2cosx

Thus, fy=2y2cosx (2)

Set the above partial derivatives to 0 and find the values of x and y.

From the equation (1),

2ysinx=0y=0,sinx=0y=0,x=0,π,2π

Substitute y=0 in the equation (2),

2(0)2cosx=02cosx=0cosx=0x=π2,3π2

Substitute x=0 in the equation (2),

2y2cos(0)=02y2(1)=02y2=0y=1

Substitute x=π in the equation (2),

2y2cos(π)=02y2(1)=02y+2=0y=1

Substitute x=2π in the equation (2),

2y2cos(2π)=02y2(1)=02y2=0y=1

Thus, the critical points are, (0,1), (π2,0) , (π,1) , (3π2,0) and (2π,1) .

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

2fx2=x(2ysinx)=2yx(sinx)=2ycosx

Hence, 2fx2=2ycosx .

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

2fy2=y(2y2cosx)=y(2y)y(2cosx)=20=2

Hence, 2fy2=2 .

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

2fxy=y(2ysinx)=2sinxy(y)=2sinx(1)=2sinx

Hence, 2fxy=2sinx .

Use second derivative test and obtain the value of D for the critical point (π2,0) .

D(π2,0)=fxx(π2,0)fyy(π2,0)[fxy(π2,0)]2=(0)(2)(2sin(π2))2=(2)2=4

Thus, D(π2,0)<0 and hence there exists a saddle point

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Chapter 14 Solutions

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P-14RECh-14 P-15RECh-14 P-16RECh-14 P-17RECh-14 P-18RECh-14 P-19RECh-14 P-20RECh-14 P-21RECh-14 P-22RECh-14 P-23RECh-14 P-24RECh-14 P-25RECh-14 P-26RECh-14 P-27RECh-14 P-28RECh-14 P-29RECh-14 P-30RECh-14 P-31RECh-14 P-32RECh-14 P-33RECh-14 P-34RECh-14 P-35RECh-14 P-36RECh-14 P-37RECh-14 P-38RECh-14 P-39RECh-14 P-40RECh-14 P-41RECh-14 P-42RECh-14 P-43RECh-14 P-44RECh-14 P-45RECh-14 P-46RECh-14 P-47RECh-14 P-48RECh-14 P-49RECh-14 P-50RECh-14 P-51RECh-14 P-52RECh-14 P-53RECh-14 P-54RECh-14 P-55RECh-14 P-56RECh-14 P-57RECh-14 P-58RECh-14 P-59RECh-14 P-60RECh-14 P-61RECh-14 P-62RECh-14 P-63RECh-14 P-64RECh-14 P-65RECh-14 P-1PCh-14 P-2PCh-14 P-3PCh-14 P-4PCh-14 P-5PCh-14 P-6PCh-14 P-7PCh-14 P-8P

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