   Chapter 14.7, Problem 6E

Chapter
Section
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.6. f(x, y) = xy − 2x − 2y – x2 – y2

To determine

To find: The local maximum, local minimum and saddle point of the

function f(x,y)=xy2x2yx2y2 .

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:

The given function is, f(x,y)=xy2x2yx2y2 .

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

fx=x(xy2x2yx2y2)=x(xy)x(2x)x(2y)x(x2)x(y2)=y202x0=y22x

Thus, fx=y22x . (1)

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

fy=y(xy2x2yx2y2)=y(xy)y(2x)y(2y)y(x2)y(y2)=x0202y=x22y

Thus, fy=x22y . (2)

Solve the equations (1) and (2) and find the values of x and y.

From the equation (1),

y22x=0y2=2xy=2x+2

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

x22(2x+2)=0x24x4=03x6=0x=2

Substitute the value of x=2 in the equation y=2x+2 and obtain y value.

y=2(2)+2=4+2=2

Thus, the critical point is, (x,y)=(2,2)

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