   Chapter 14.7, Problem 4E

Chapter
Section
Textbook Problem

Use the level curves in the figure to predict the location of the critical points of f and whether f has a saddle point or a local maximum or minimum at each critical point. Explain your reasoning. Then use the Second Derivatives Test to confirm your predictions.4. f(x, y) = 3x − x3 − 2y2 +3 y4 To determine

To predict: The location of the critical points of f and whether f has a local maximum, local minimum and saddle point at each critical point using the level curves in the figure. Use second derivative test to confirm the predictions.

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:

From the given figure it is clear that the points (1,1),(1,1) are surrounded by an oval shape.

Also, it can be identified that the value of f is increasing around all the points of (1,±1) .

Thus, the local minimum occurs at (1,±1) or near the point (1,±1) . Therefore, the critical point is (1,±1) .

The function is decreasing at the point (1,0) and hence it is local maximum

Moreover observe that near the points (1,0),(1,1),(1,1) , the values of f is increasing in some directions and decreasing in some other direction. So, there exists the saddle points at (1,0),(1,1),(1,1) .

Verification:

The given function is, f(x,y)=3xx32y2+y4 .

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

fx=x(3xx32y2+y4)=x(3x)x(x3)x(2y2)+x(y4)=3(1)3x20+0=33x2

Thus, fx=33x2 . (1)

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

fy=y(3xx32y2+y4)=y(3x)y(x3)y(2y2)+y(y4)=002(2y)+4y3=4y+4y3

Thus, fy=4y+4y3 . (2)

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

From the equation (1),

33x2=03x2=3x2=1x=±1

From the equation (2),

4y+4y3=04y(y21)=0y=0,y2=1y=0,y=±1

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

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

2fx2=x(33x2)=x(3)x(3x2)=03(2x)=6x

Hence, 2fx2=6x .

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

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

Hence, 2fy2=4+12y2 .

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

2fxy=y(33x2)=y(3)y(3x2)=00=0

Hence, 2fxy=0

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