What is the extreme f(x,y)=x^2y+12x^2+(3y^2)/2+5 ?

The extreme f(x,y)=x^2y+12x^2+(3y^2)/2+5 is Minimum(0,0),Saddle(6,-12),Saddle(-6,-12)

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extreme f(x,y)=x^2y+12x^2+(3y^2)/2+5

extreme points

f (x, y) = x^{2} y + 12 x^{2} + \frac{3 y ^{2}}{2} + 5

Minimum (0, 0), Saddle (6, - 12), Saddle (- 6, - 12)

Solution steps

Find the critical points:

(0, 0), (6, - 12), (- 6, - 12)

\frac{\partial ^{2} f}{\partial x ^{2}} = 3

D (x, y) = 3 (2 y + 24) - 4 x^{2}

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Check critical point

(6, - 12) :

Saddle

Check critical point

(- 6, - 12) :

Saddle

Minimum (0, 0), Saddle (6, - 12), Saddle (- 6, - 12)

f (x) = x^{\frac{2}{3}}, - 1 \leq x \leq 27

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

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

y = \frac{x}{x ^{2} + 4}

y = x (x - 1)^{5}

What is the extreme f(x,y)=x^2y+12x^2+(3y^2)/2+5 ?
The extreme f(x,y)=x^2y+12x^2+(3y^2)/2+5 is Minimum(0,0),Saddle(6,-12),Saddle(-6,-12)