What is the extreme f(x,y)=10x^2y-5x^2-4y^2-x^4-2y^4 ?

The extreme f(x,y)=10x^2y-5x^2-4y^2-x^4-2y^4 is

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extreme f(x,y)=10x^2y-5x^2-4y^2-x^4-2y^4

extreme points

f (x, y) = 10 x^{2} y - 5 x^{2} - 4 y^{2} - x^{4} - 2 y^{4}

Saddle (\frac{7.33861 E 15}{1.0 E 16}, \frac{8.08465 E 15}{1.25 E 16}), Saddle (- \frac{7.33861 E 15}{1.0 E 16}, \frac{8.08465 E 15}{1.25 E 16}), Maximum (\frac{6.99192 E 15}{1.0 E 15}, \frac{1.18649 E 15}{6.25 E 14}), Maximum (- \frac{6.99192 E 15}{1.0 E 15}, \frac{1.18649 E 15}{6.25 E 14})

Solution steps

Find the critical points:

(\frac{7.33861 E 15}{1.0 E 16}, \frac{8.08465 E 15}{1.25 E 16}), (- \frac{7.33861 E 15}{1.0 E 16}, \frac{8.08465 E 15}{1.25 E 16}), (\frac{6.99192 E 15}{1.0 E 15}, \frac{1.18649 E 15}{6.25 E 14}), (- \frac{6.99192 E 15}{1.0 E 15}, \frac{1.18649 E 15}{6.25 E 14})

\frac{\partial ^{2} f}{\partial x ^{2}} = - 24 y^{2} - 8

D (x, y) = - 480 y^{3} + 288 x^{2} y^{2} + 240 y^{2} - 160 y - 304 x^{2} + 80

Check the sign of at each critical point

Check critical point

(\frac{7.33861 E 15}{1.0 E 16}, \frac{8.08465 E 15}{1.25 E 16}) :

Saddle

Check critical point

(- \frac{7.33861 E 15}{1.0 E 16}, \frac{8.08465 E 15}{1.25 E 16}) :

Saddle

Check critical point

(\frac{6.99192 E 15}{1.0 E 15}, \frac{1.18649 E 15}{6.25 E 14}) :

Maximum

Check critical point

(- \frac{6.99192 E 15}{1.0 E 15}, \frac{1.18649 E 15}{6.25 E 14}) :

Maximum

Saddle (\frac{7.33861 E 15}{1.0 E 16}, \frac{8.08465 E 15}{1.25 E 16}), Saddle (- \frac{7.33861 E 15}{1.0 E 16}, \frac{8.08465 E 15}{1.25 E 16}), Maximum (\frac{6.99192 E 15}{1.0 E 15}, \frac{1.18649 E 15}{6.25 E 14}), Maximum (- \frac{6.99192 E 15}{1.0 E 15}, \frac{1.18649 E 15}{6.25 E 14})

f (x) = x^{\frac{4}{5}} (x - 9)^{2}

\frac{1}{4} x + 3

f (x, y) = y e^{x}

f (x) = x^{3} + 18 x^{2} + 81 x

f (x) = x^{4} - 3 x^{3} + 3 x^{2} + 1

What is the extreme f(x,y)=10x^2y-5x^2-4y^2-x^4-2y^4 ?
The extreme f(x,y)=10x^2y-5x^2-4y^2-x^4-2y^4 is