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
