What is the extreme f(x,y)=2x^3+4xy-y^2 ?

The extreme f(x,y)=2x^3+4xy-y^2 is Saddle(0,0),Maximum(-4/3 ,-8/3)

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extreme f(x,y)=2x^3+4xy-y^2

extreme points

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

Saddle (0, 0), Maximum (- \frac{4}{3}, - \frac{8}{3})

Solution steps

Find the critical points:

(0, 0), (- \frac{4}{3}, - \frac{8}{3})

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

D (x, y) = - 24 x - 16

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(- \frac{4}{3}, - \frac{8}{3}) :

Maximum

Saddle (0, 0), Maximum (- \frac{4}{3}, - \frac{8}{3})

f (x, y) = x^{3} + y^{3} - 36 x y

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

f (x) = x^{2} - 2 x + 1, 0 \leq x \leq 3

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

3 x^{2} - 24 ln (x)

What is the extreme f(x,y)=2x^3+4xy-y^2 ?
The extreme f(x,y)=2x^3+4xy-y^2 is Saddle(0,0),Maximum(-4/3 ,-8/3)