What is the extreme f(x,y)=-5y^2-6x^3+12x^2 ?

The extreme f(x,y)=-5y^2-6x^3+12x^2 is Saddle(0,0),Maximum(4/3 ,0)

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extreme f(x,y)=-5y^2-6x^3+12x^2

extreme points

f (x, y) = - 5 y^{2} - 6 x^{3} + 12 x^{2}

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

Solution steps

Find the critical points:

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

\frac{\partial ^{2} f}{\partial x ^{2}} = - 10

D (x, y) = - 10 (- 36 x + 24)

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

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

Maximum

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

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

f (x) = 4 x^{2} - 16 x + 16

f (x) = \frac{4}{4 - x}

f (x) = \frac{( x - 6 )}{x ^{2}}

f (x) = (x - 2)^{3} + 8

What is the extreme f(x,y)=-5y^2-6x^3+12x^2 ?
The extreme f(x,y)=-5y^2-6x^3+12x^2 is Saddle(0,0),Maximum(4/3 ,0)