Question

Question #64e5b

Answer 1

`f(x,y)=2x^3y^2-x^4y^2-x^3y^3`
assuming y as constant
`d(f(x,y))/dx=6x^2y^2-4x^3y^2-3x^2y^3=f_x`
assuming x as constant
`d(f(x,y))/dy=4x^3y-2x^4y-3x^3y^2=f_y`
for critical points partial derrivatives must be equal to 0
#f_x=0=> 6x^2y^2-4x^3y^2-3x^2y^3=0 =>(xy)^2(6-4x-3y)=0#
#f_y=0=>4x^3y-2x^4y-3x^3y^2 =>x^3y(4-2x-3y)=0#
one of the critical point is xy= 0 ie every point on the x y axis
or
`6-4x-3y=0,4-2x-3y=0`
#=> 6-4x=4-2x[=>2x=2=>x=1 =>3y=2=>y=2/3#
the critical points are `xy=0 and (1,2/3)`