Chapter 14, Problem 57RE

### Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

Chapter
Section

### Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550
Textbook Problem

# Use a graph or level curves or both to estimate the local maximum and minimum values and saddle points of f(x, y) = x3− 3x + y4 − 2y2. Then use calculus to find these values precisely.

To determine

To find: The local maximum, local minimum and saddle point using a graph or level curves and use calculus to find these values for the function f(x,y)=x33x+y42y2 .

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)=x33x+y42y2 .

Calculation:

Use online graphing calculator and obtain the graph of f(x,y)=x33x+y42y2 as shown below in Figure 1.

From Figure 1, it can be identified that there occurs a local maximum at the point (1,0) , the local minimum at the point (1,±1) and the saddle points at (1,0),(1,±1) .

Find the exact values of local maximum, local minimum and saddle point using calculus as follows.

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

fx=x(x33x+y42y2)=x(x3)x(3x)+x(y4)x(2y2)=3x23(1)+00=3x23

Thus, fx=3x23 (1)

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

fy=y(x33x+y42y2)=y(x3)y(3x)+y(y4)y(2y2)=00+4y32(2y)=4y34y

Thus, fy=4y34y (2)

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

From the equation (1),

3x23=03x2=3x2=1x=±1

From the equation (2),

4y34y=04y34y=04y(y21)=0y=0,y=±1

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

Obtain the second derivatives as follows.

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

2fx2=x(3x23)=x(3x2)x(3)=3(2x)0=6x

Hence, 2fx2=6x .

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

2fy2=y(4y34y)=y(4y3)y(4y)=4(3y2)4(1)=12y24

Hence, 2fy2=12y24

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