   Chapter 14.3, Problem 14E

Chapter
Section
Textbook Problem

Find fx and fy and graph f, fx, and fy with domains and viewpoints that enable you to see the relationships between them.14. f ( x , y ) = y 1 + x 2 y 2

To determine

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

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”.

Calculation:

The given function is, f(x,y)=ycosx .

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

fx=x(ycosx)=yx(cosx)=y(sinx)=ysinx

Thus, fx=ysinx (1)

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

fy=y(ycosx)=cosxy(y)=cosx(1)=cosx

Thus, fy=cosx (2)

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

From the equation (1),

ysinx=0y=0,sinx0

From the equation (2),

cosx=0

The values of both cosx and sinx cannot be zero for the same x value. Thus, it is not possible.

So, take x=π2+nπ where n is an integer, n .

Thus, the critical points is, (x,y)=(π2+nπ,0) .

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

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