# Question 64e5b

Sep 21, 2015

$f \left(x , y\right) = 2 {x}^{3} {y}^{2} - {x}^{4} {y}^{2} - {x}^{3} {y}^{3}$
assuming y as constant
$d \frac{f \left(x , y\right)}{\mathrm{dx}} = 6 {x}^{2} {y}^{2} - 4 {x}^{3} {y}^{2} - 3 {x}^{2} {y}^{3} = {f}_{x}$
assuming x as constant
$d \frac{f \left(x , y\right)}{\mathrm{dy}} = 4 {x}^{3} y - 2 {x}^{4} y - 3 {x}^{3} {y}^{2} = {f}_{y}$
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
$6 - 4 x - 3 y = 0 , 4 - 2 x - 3 y = 0$
=> 6-4x=4-2x[=>2x=2=>x=1 =>3y=2=>y=2/3#
the critical points are $x y = 0 \mathmr{and} \left(1 , \frac{2}{3}\right)$