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

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

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

extreme points

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

Saddle (- 1, - 1), Maximum (1, - 1)

Solution steps

Find the critical points:

(- 1, - 1), (1, - 1)

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

D (x, y) = 12 x

Check the sign of at each critical point

Check critical point

(- 1, - 1) :

Saddle

Check critical point

(1, - 1) :

Maximum

Saddle (- 1, - 1), Maximum (1, - 1)

\frac{9}{8} y \frac{3}{4}

f (x) = - 2 x^{3} + 27 x^{2} - 108 x, 2 \leq x \leq 7

f (x, y) = 64 - 4 x^{2} - y^{2}

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

\frac{- 8 x + 75}{9 x - 61}

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