What is the extreme f(x,y)=2x^3+2y^2-800x ?

The extreme f(x,y)=2x^3+2y^2-800x is Saddle(-(20)/(sqrt(3)),0),Minimum((20)/(sqrt(3)),0)

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extreme f(x,y)=2x^3+2y^2-800x

extreme points

f (x, y) = 2 x^{3} + 2 y^{2} - 800 x

Saddle (- \frac{20}{3}, 0), Minimum (\frac{20}{3}, 0)

Solution steps

Find the critical points:

(- \frac{20}{3}, 0), (\frac{20}{3}, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 4

D (x, y) = 48 x

Check the sign of at each critical point

Check critical point

(- \frac{20}{3}, 0) :

Saddle

Check critical point

(\frac{20}{3}, 0) :

Minimum

Saddle (- \frac{20}{3}, 0), Minimum (\frac{20}{3}, 0)

f (x) = 3 x^{3} - 4 x^{2} - 6 x + 1

f (x, y) = - 6 x^{2} - y^{2} + 9 x - 5 y + 8

f (x) = x^{2} + \frac{240}{x}

f (x) = 6 x^{4} - 8 x^{3} - 24 x^{2} + 35

f (x) = x + \frac{13}{x}

What is the extreme f(x,y)=2x^3+2y^2-800x ?
The extreme f(x,y)=2x^3+2y^2-800x is Saddle(-(20)/(sqrt(3)),0),Minimum((20)/(sqrt(3)),0)