Chapter 14, Problem 53RE

Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

Chapter
Section

Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550
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.53. f(x, y) = 3xy – x2y – xy2

To determine

To find: The local maximum, local minimum and saddle point of the function, f(x,y)=3xyx2yxy2 .

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)=3xyx2yxy2 .

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

fx=x(3xyx2yxy2)=x(3xy)x(x2y)x(xy2)=3y(1)y(2x)y2(1)=3y2xyy2

Thus, fx=3y2xyy2 (1)

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

fy=y(3xyx2yxy2)=y(3xy)y(x2y)y(xy2)=3x(1)x2(1)x(2y)=3xx22xy

Thus, fy=3xx22xy (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),

3y2xyy2=0y(32xy)=0y=0,y=32x

Substitute y=32x in the equation (2) and obtain the value of x.

3xx22x(32x)=03xx26x+4x2=03x2=3xx=1

Substitute x=1 in the equation y=32x and obtain y value.

y=32(1)=32=1

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

3xx22x(0)=03xx2=0x(3x)=0x=0,x=3

Substitute x=0 in the equation y=32x and obtain y value.

y=32(0)=30=3

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

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

2fx2=x(3y2xyy2)=x(3y)x(2xy)x(y2)=02y(1)0=2y

Hence, 2fx2=2y .

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

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