Question

A. Find the critical point of the given function and then determine whether it is a local maximum, local minimum, or saddle point.
f(x, y) = x2 − y2 + 3xy

b. Find the critical point of the given function and then determine whether it is a local maximum, local minimum, or saddle point.

f(x, y) = x2 + y2 − xy

Answer 1

Answer:

Step-by-step explanation:

a) f(x,y) =  x² - y² + 3xy

First derivative x (variable)                          First derivative y (variable)

f´(x)  = 2x  + 3y                                                      f´(y)  = - 2y + 3x

Second derivative x (variable)                   Second derivative y (variable)

f´´(xx) = 2                                                               f´´ (yy) = -2

Cross derivative                                              Cross  derivative

f´´(x,y )  =   3                                                         f´´(y,x)  =  3

f´(x) = 0                                  and                     f´(y)  =  0

2x  + 3y  = 0                                                    - 2y + 3x  = 0

We got a two equation system with two uknown variables

2x  + 3y  = 0             - 2y + 3x  = 0

Solving

x  =  3/2 * y                 -2y  +  3* (3/2) *y =0

-2y  +  9/2 y  = 0           y  = 0     and    x  =  0  

Critical point  P ( 0 ,0 )

we must evaluate

f´´(x,x)  = 2

f´´(y,y)  = -2

f´´ (x,y)   = f´´(y,x) = 3

We compute discriminiting

D = f´´(x,x) * f´´(y,y) - [f´´(x,y)]²         D = (2)*(-2)  - (3)²     D = -4 -9

D  = -13

Then we got one second derivative positive the other negative and D < 0

we have a saddle point

b)  f(x,y)  =  x²   + y²  - xy

folowing the same procedure

f¨(x)    =  2x -y                                          f´(y)     = 2y - x

f´´ (x,x) = 2                                                 f´´(y,y) = 2

Croos derivative

f´´(x,y) = -1                                             f´´(y,x)  =   -1

Equation system:    

2x -y   = 0                  y  = 2x

2y - x   = 0            2(2x)  - x  = 0        4x  - x  = 0      x = 0  and y = 0

Critical point  P ( 0 , 0 )

f´´(x,x)  =  2

f´´(y,y)  =  2

f´´ (x,y) = f´´(y,x)  = -1

D = (2)*2  - (-1)     =  4 - 1 = 3

D = 3

The fuction has a minimum the point P ( 0 , 0) is a minimum