What is the extreme f(x,y)=-1/4 x^4+2/3 x^3+4xy-y^2 ?

The extreme f(x,y)=-1/4 x^4+2/3 x^3+4xy-y^2 is Saddle(0,0),Maximum(4,8),Maximum(-2,-4)

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extreme f(x,y)=-1/4 x^4+2/3 x^3+4xy-y^2

extreme points

f (x, y) = - \frac{1}{4} x^{4} + \frac{2}{3} x^{3} + 4 x y - y^{2}

Saddle (0, 0), Maximum (4, 8), Maximum (- 2, - 4)

Solution steps

Find the critical points:

(0, 0), (4, 8), (- 2, - 4)

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

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

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(4, 8) :

Maximum

Check critical point

(- 2, - 4) :

Maximum

Saddle (0, 0), Maximum (4, 8), Maximum (- 2, - 4)

f (x, y) = x y (10 - 2 x - 3 y)

P (b, s) = b + 2 s

f (x) = 2 x e^{- x}

x^{\frac{2}{3}} (\frac{5}{2} - x)

f (x) = 2 x^{5} + 5 x^{4}

What is the extreme f(x,y)=-1/4 x^4+2/3 x^3+4xy-y^2 ?
The extreme f(x,y)=-1/4 x^4+2/3 x^3+4xy-y^2 is Saddle(0,0),Maximum(4,8),Maximum(-2,-4)