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

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

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

extreme points

f (x) = 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, y) = \frac{23}{2} y^{4} - 46 x y + 23 x^{2}

- 8 x^{\frac{5}{3}} + 3 x

f (x, y) = 2 x^{3} + 6 x y + 3 y^{2}

f (x, y) = 16 x 3 y - 2 x^{2} - 4 x - 16 y + 2

f (x) = cos (5 x), - \frac{π}{6} \leq x \leq \frac{π}{4}

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