   Chapter 14, Problem 54RE

Chapter
Section
Textbook Problem

Find the local maximum and minimum values and saddle points 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.54. f(x, y) = (x2 + y)ey/2

To determine

To find: The local maximum, local minimum and saddle point of the function, f(x,y)=(x2+y)ey2 .

Explanation

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)=(x2+y)ey2 .

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

fx=x((x2+y)ey2)=ey2x(x2+y)=ey2(2x+0)=2xey2

Thus, fx=2xey2 (1)

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

fy=y((x2+y)ey2)=[ey2(0+1)+(x2+y)ey2(12)]=ey2+12ey2(x2+y)

Thus, fy=ey2+12ey2(x2+y) (2)

Set the above derivative to 0 and solve the equations (1) and (2) to get the values of x and y.

From the equation (1),

2xey2=02x=0x=0

Substitute x=0 in the equation (2) and obtain the value of y.

ey2+12ey2(02+y)=0ey2+12yey2=012yey2=ey2y=2

Thus, the critical point is, (0,2) .

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

2fx2=x(2xey2)=ey2x(2x)=ey2(2)=2ey2

Hence, 2fx2=2ey2 .

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

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