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

The extreme f(x,y)=3x^2+2y^2-4y is Minimum(0,1)

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extreme f(x,y)=3x^2+2y^2-4y

extreme points

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

Minimum (0, 1)

Solution steps

Find the critical points:

(0, 1)

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

D (x, y) = 24

Check the sign of at each critical point

Check critical point

(0, 1) :

Minimum

Minimum (0, 1)

f (x) = \frac{3}{7} (x^{2} - 25)^{\frac{2}{3}}

- 7 x^{2} + 14 x + 245

f (x) = 15 x + 20 y

f (x) = x^{3} - 6 x^{2} + 12 x + 15

f x 4 - x

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