What is the extreme f(x,y)=10xy-x^3-5y^2 ?

The extreme f(x,y)=10xy-x^3-5y^2 is Saddle(0,0),Maximum(10/3 , 10/3)

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extreme f(x,y)=10xy-x^3-5y^2

extreme points

f (x, y) = 10 x y - x^{3} - 5 y^{2}

Saddle (0, 0), Maximum (\frac{10}{3}, \frac{10}{3})

Solution steps

Find the critical points:

(0, 0), (\frac{10}{3}, \frac{10}{3})

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

D (x, y) = 60 x - 100

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(\frac{10}{3}, \frac{10}{3}) :

Maximum

Saddle (0, 0), Maximum (\frac{10}{3}, \frac{10}{3})

f (x) = ln (7 - 2 x^{2})

f (x) = \frac{x ^{3}}{3} - \frac{x ^{2}}{2} - 6 x + 4

\frac{x}{x ^{2} + 13 x + 36}

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

f (x) = 2 x^{3} - 12 x^{2} + 4

What is the extreme f(x,y)=10xy-x^3-5y^2 ?
The extreme f(x,y)=10xy-x^3-5y^2 is Saddle(0,0),Maximum(10/3 , 10/3)