   Chapter 14, Problem 52RE

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.52. f(x, y) = x3 − 6xy + 8y3

To determine

To find: The local maximum, local minimum and saddle point of the function, f(x,y)=x36xy+8y3 .

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)=x36xy+8y3 .

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

fx=x(x36xy+8y3)=x(x3)x(6xy)+x(8y3)=3x26y(1)+0=3x26y

Thus, fx=3x26y (1)

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

fy=y(x36xy+8y3)=y(x3)y(6xy)+y(8y3)=06x(1)+8(3y2)=6x+24y2

Thus, fy=6x+24y2 (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),

3x26y=03x2=6yx2=2yy=x22

Substitute y=x22 in the equation (2) and obtain the value of x.

6x+24(x22)2=06x+24(x44)=06x+6x4=06x(1x3)=0

Simplify further as follows.

x=0,x3=1x=0,x=1

Substitute x=0 in the equation y=x22 and obtain y value.

y=(0)22y=0

Substitute x=1 in the equation y=x22 and obtain y value.

y=(1)22y=12

Thus, the critical points are, (0,0) and (1,12) .

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

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