What is the extreme f(x,y)= 1/3 x^3-3x^2+(y^2)/4+xy+13x-y+2 ?

The extreme f(x,y)= 1/3 x^3-3x^2+(y^2)/4+xy+13x-y+2 is Saddle(3,-4),Minimum(5,-8)

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extreme f(x,y)= 1/3 x^3-3x^2+(y^2)/4+xy+13x-y+2

extreme points

f (x, y) = \frac{1}{3} x^{3} - 3 x^{2} + \frac{y ^{2}}{4} + x y + 13 x - y + 2

Saddle (3, - 4), Minimum (5, - 8)

Solution steps

Find the critical points:

(3, - 4), (5, - 8)

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{1}{2}

D (x, y) = x - 4

Check the sign of at each critical point

Check critical point

(3, - 4) :

Saddle

Check critical point

(5, - 8) :

Minimum

Saddle (3, - 4), Minimum (5, - 8)

\frac{x ^{2}}{x + 1}

x^{\frac{1}{3}} (x + 3)^{\frac{2}{3}}

f (x) = 6 x^{2} - 24 x + 18

f (x) = x (17 - 45 + 2 x) (\frac{45}{2} - x)

f (x) = x^{4} (x - 3)^{3}

What is the extreme f(x,y)= 1/3 x^3-3x^2+(y^2)/4+xy+13x-y+2 ?
The extreme f(x,y)= 1/3 x^3-3x^2+(y^2)/4+xy+13x-y+2 is Saddle(3,-4),Minimum(5,-8)