What is the extreme f(x,y)=4x^2+9y^2-2x-18xy^2 ?

The extreme f(x,y)=4x^2+9y^2-2x-18xy^2 is Minimum(1/4 ,0),Saddle(1/2 , 1/3),Saddle(1/2 ,-1/3)

extreme f(x,y)=4x^2+9y^2-2x-18xy^2

extreme points

f (x, y) = 4 x^{2} + 9 y^{2} - 2 x - 18 x y^{2}

Minimum (\frac{1}{4}, 0), Saddle (\frac{1}{2}, \frac{1}{3}), Saddle (\frac{1}{2}, - \frac{1}{3})

Solution steps

Find the critical points:

(\frac{1}{4}, 0), (\frac{1}{2}, \frac{1}{3}), (\frac{1}{2}, - \frac{1}{3})

\frac{\partial ^{2} f}{\partial x ^{2}} = 18 - 36 x

D (x, y) = 8 (18 - 36 x) - 1296 y^{2}

Check the sign of at each critical point

Check critical point

(\frac{1}{4}, 0) :

Minimum

Check critical point

(\frac{1}{2}, \frac{1}{3}) :

Saddle

Check critical point

(\frac{1}{2}, - \frac{1}{3}) :

Saddle

Minimum (\frac{1}{4}, 0), Saddle (\frac{1}{2}, \frac{1}{3}), Saddle (\frac{1}{2}, - \frac{1}{3})

f (x) = \frac{x ^{3}}{3} - 10 x^{2} + 96 x + 100

f (x) = - x^{3} + 9 x^{2} + 120 x - 200, x \geq 5

f (x) = x^{\frac{3}{4}} - 2 x^{\frac{3}{4}}

f (x) = x^{\frac{2}{3}} (x - 6)

\frac{16 a ^{4} + 16 a}{a ^{2} + 1 - a}

What is the extreme f(x,y)=4x^2+9y^2-2x-18xy^2 ?
The extreme f(x,y)=4x^2+9y^2-2x-18xy^2 is Minimum(1/4 ,0),Saddle(1/2 , 1/3),Saddle(1/2 ,-1/3)