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.

12. f(x, y) = x³ + y³ − 3x² − 3y² − 9x

To find: The local maximum, local minimum and saddle point of the function f(x,y)=x3+y3−3x2−3y2−9x .

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)=x3+y3−3x2−3y2−9x .

Calculation:

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

∂f∂x=∂∂x(x3+y3−3x2−3y2−9x)=∂∂x(x3)+∂∂x(y3)−3∂∂x(x2)−3∂∂x(y2)−9∂∂x(x)=3x2+0−3(2x)−0−9(1)=3x2−6x−9

Thus, ∂f∂x=3x2−6x−9 . (1)

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

∂f∂y=∂∂y(x3+y3−3x2−3y2−9x)=∂∂y(x3)+∂∂y(y3)−∂∂y(3x2)−∂∂y(3y2)−9∂∂y(x)=0+3y2−0−3(2y)−0=3y2−6y

Thus, ∂f∂y=3y2−6y (2)

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

From the equation (2),

3y2−6y=03y(y−2)=03y=0,y−2=0y=0,y=2

From the equation (1),

3x2−6x−9=03(x2−2x−3)=03(x+1)(x−3)=0x=−1,3

Thus, the critical points are, (−1,0) , (−1,2) , (3,0) and (3,2) .

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

∂2f∂x2=∂∂x(3x2−6x−9)=3∂∂x(x2)−6∂∂x(x)−∂∂x(9)=3(2x)−6(1)−0=6x−6

Hence, ∂2f∂x2=6x−6 .

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

∂2f∂y2=∂∂y(3y2−6y)=3∂∂y(y2)−6∂∂y(y)=3(2y)−6(1)=6y−6

Hence, ∂2f∂y2=6y−6 .

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

∂2f∂x∂y=∂∂y(3x2−6x−9)=∂∂y(3x2)−∂∂y(6x)−∂∂y(9)=0

Hence, ∂2f∂x∂y=0