What is the extreme f(x,y)=-4y^2-5x^3+13x^2 ?

The extreme f(x,y)=-4y^2-5x^3+13x^2 is Saddle(0,0),Maximum(26/15 ,0)

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extreme f(x,y)=-4y^2-5x^3+13x^2

extreme points

f (x, y) = - 4 y^{2} - 5 x^{3} + 13 x^{2}

Saddle (0, 0), Maximum (\frac{26}{15}, 0)

Solution steps

Find the critical points:

(0, 0), (\frac{26}{15}, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 8

D (x, y) = - 8 (- 30 x + 26)

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(\frac{26}{15}, 0) :

Maximum

Saddle (0, 0), Maximum (\frac{26}{15}, 0)

f (x, y) = x - \frac{1}{2} y^{2} - \frac{1}{3} x^{3} + y + 6

f (x) = \frac{4 x ^{2}}{2 x - 8}

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

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

f (x) = 8 x^{2} - 4 x^{4}

What is the extreme f(x,y)=-4y^2-5x^3+13x^2 ?
The extreme f(x,y)=-4y^2-5x^3+13x^2 is Saddle(0,0),Maximum(26/15 ,0)