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

The extreme f(x,y)=(x^3)/3-3y^2-(x^2)/2+3xy is Saddle(0,0),Maximum(-1/2 ,-1/4)

Popular >

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

extreme points

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

Saddle (0, 0), Maximum (- \frac{1}{2}, - \frac{1}{4})

Solution steps

Find the critical points:

(0, 0), (- \frac{1}{2}, - \frac{1}{4})

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

D (x, y) = - 12 x - 3

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(- \frac{1}{2}, - \frac{1}{4}) :

Maximum

Saddle (0, 0), Maximum (- \frac{1}{2}, - \frac{1}{4})

y = \frac{( x + 3 ) ( x ^{2} - 2 r )}{x ^{3}}

14 x^{2} + \frac{3500}{x}

f (x, y) = x^{3} + y^{3} - 12 x - 3 y + 15

f (x) = x^{3} + 1 5^{2}

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

What is the extreme f(x,y)=(x^3)/3-3y^2-(x^2)/2+3xy ?
The extreme f(x,y)=(x^3)/3-3y^2-(x^2)/2+3xy is Saddle(0,0),Maximum(-1/2 ,-1/4)