What is the extreme f(x,y)=3x^2+y^3-18xy+22 ?

The extreme f(x,y)=3x^2+y^3-18xy+22 is Minimum(54,18),Saddle(0,0)

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extreme f(x,y)=3x^2+y^3-18xy+22

extreme points

f (x, y) = 3 x^{2} + y^{3} - 18 x y + 22

Minimum (54, 18), Saddle (0, 0)

Solution steps

Find the critical points:

(54, 18), (0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 6 y

D (x, y) = 36 y - 324

Check the sign of at each critical point

Check critical point

(54, 18) :

Minimum

Check critical point

(0, 0) :

Saddle

Minimum (54, 18), Saddle (0, 0)

f (3.2) = - \frac{40000 t}{( 3 + t ^{2} + 2 x ^{2} ) ^{2}}

\frac{4 e ^{- 2 x}}{2 x + 5}

f (x, y) = - 2 x^{2} - y^{3} + 9 y^{2} + 16 x - 15 y + 5

f (x) = 2 x^{2} + \frac{91.2}{x}

P (a) = y^{2} - 4 r^{2} + r + e^{2}

What is the extreme f(x,y)=3x^2+y^3-18xy+22 ?
The extreme f(x,y)=3x^2+y^3-18xy+22 is Minimum(54,18),Saddle(0,0)