What is the extreme-x^3+4xy-2y^2+1 ?

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

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extreme-x^3+4xy-2y^2+1

extreme points

- x^{3} + 4 x y - 2 y^{2} + 1

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

Solution steps

Find the critical points:

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

\frac{\partial ^{2} f}{\partial x ^{2}} = - 4

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{4}{3}) :

Maximum

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

f (x, y) = x^{2} + 2 y^{2} - x y + 14 y

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

f (x) = - x^{3} + 12 x, (- 6, 5)

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

f (x, y) = 8 y^{2} + x^{2} - x^{2} y

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