What is the extreme f(x,y)=9x^2+9xy+6y^2+4xy^2+6x^2y ?

The extreme f(x,y)=9x^2+9xy+6y^2+4xy^2+6x^2y is

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extreme f(x,y)=9x^2+9xy+6y^2+4xy^2+6x^2y

extreme points

f (x, y) = 9 x^{2} + 9 x y + 6 y^{2} + 4 x y^{2} + 6 x^{2} y

Minimum (0, 0), Saddle (\frac{5}{3}, - \frac{5}{4}), Saddle (- \frac{3}{2}, \frac{9 + 3 57}{8}), Saddle (- \frac{3}{2}, \frac{9 - 3 57}{8})

Solution steps

Find the critical points:

(0, 0), (\frac{5}{3}, - \frac{5}{4}), (- \frac{3}{2}, \frac{9 + 3 57}{8}), (- \frac{3}{2}, \frac{9 - 3 57}{8})

\frac{\partial ^{2} f}{\partial x ^{2}} = 8 x + 12

D (x, y) = - 144 x^{2} - 96 x y - 72 x + 135 - 64 y^{2}

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Check critical point

(\frac{5}{3}, - \frac{5}{4}) :

Saddle

Check critical point

(- \frac{3}{2}, \frac{9 + 3 57}{8}) :

Saddle

Check critical point

(- \frac{3}{2}, \frac{9 - 3 57}{8}) :

Saddle

Minimum (0, 0), Saddle (\frac{5}{3}, - \frac{5}{4}), Saddle (- \frac{3}{2}, \frac{9 + 3 57}{8}), Saddle (- \frac{3}{2}, \frac{9 - 3 57}{8})

f (x) = x^{\frac{6}{7}}, - 2 \leq x \leq 4

f (x, y) = \frac{1}{e ^{x - y}}

f (x, y) = \frac{x ^{3}}{3} - 3 y^{2} - \frac{x ^{2}}{2} + 3 x y

y = \frac{( x + 3 ) ( x ^{2} - 2 r )}{x ^{3}}

14 x^{2} + \frac{3500}{x}

What is the extreme f(x,y)=9x^2+9xy+6y^2+4xy^2+6x^2y ?
The extreme f(x,y)=9x^2+9xy+6y^2+4xy^2+6x^2y is