   Chapter 14, Problem 51RE

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.51. f(x, y) = x2− xy + y2 + 9x – 6y + 10

To determine

To find: The local maximum, local minimum and saddle point of the function, f(x,y)=x2xy+y2+9x6y+10 .

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)=x2xy+y2+9x6y+10 .

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

fx=x(x2xy+y2+9x6y+10)=x(x2)x(xy)+x(y2)+x(9x)x(6y)+x(10)=2xy(1)+0+9(1)0+0=2xy+9

Thus, fx=2xy+9 (1)

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

fy=y(x2xy+y2+9x6y+10)=y(x2)y(xy)+y(y2)+y(9x)y(6y)+y(10)=0x(1)+2y+06(1)+0=x+2y6

Thus, fy=x+2y6 (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),

2xy+9=02x+9=y

Substitute y=2x+9 in the equation (2) and obtain the value of x.

x+2(2x+9)6=0x+4x+186=03x=12x=4

Substitute x=4 in the equation y=2x+9 and obtain y value

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