Chapter 14.7, Problem 12E

### Multivariable Calculus

8th Edition
James Stewart
ISBN: 9781305266643

Chapter
Section

### 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.12. f(x, y) = x3 + y3 − 3x2 − 3y2 − 9x

To determine

To find: The local maximum, local minimum and saddle point of the function f(x,y)=x3+y33x23y29x .

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”.

Given:

The function is, f(x,y)=x3+y33x23y29x .

Calculation:

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

fx=x(x3+y33x23y29x)=x(x3)+x(y3)3x(x2)3x(y2)9x(x)=3x2+03(2x)09(1)=3x26x9

Thus, fx=3x26x9 . (1)

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

fy=y(x3+y33x23y29x)=y(x3)+y(y3)y(3x2)y(3y2)9y(x)=0+3y203(2y)0=3y26y

Thus, fy=3y26y (2)

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

From the equation (2),

3y26y=03y(y2)=03y=0,y2=0y=0,y=2

From the equation (1),

3x26x9=03(x22x3)=03(x+1)(x3)=0x=1,3

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

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

2fx2=x(3x26x9)=3x(x2)6x(x)x(9)=3(2x)6(1)0=6x6

Hence, 2fx2=6x6 .

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

2fy2=y(3y26y)=3y(y2)6y(y)=3(2y)6(1)=6y6

Hence, 2fy2=6y6 .

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

2fxy=y(3x26x9)=y(3x2)y(6x)y(9)=0

Hence, 2fxy=0

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