What is the extreme f(x,y)=3x^2+2y^2-4y+10y ?

The extreme f(x,y)=3x^2+2y^2-4y+10y is Minimum(0,-3/2)

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extreme f(x,y)=3x^2+2y^2-4y+10y

extreme points

f (x, y) = 3 x^{2} + 2 y^{2} - 4 y + 10 y

Minimum (0, - \frac{3}{2})

Solution steps

Find the critical points:

(0, - \frac{3}{2})

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

D (x, y) = 24

Check the sign of at each critical point

Check critical point

(0, - \frac{3}{2}) :

Minimum

Minimum (0, - \frac{3}{2})

E (a, b) = (a + b)^{2} - (a - b)^{2}

g (x) = x^{2} - 10 x + 27

f (x) = 2 - 4 (sin^{2} (x))

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

f (x) = \frac{x ^{2} ( x - 3 )}{x + 6}

What is the extreme f(x,y)=3x^2+2y^2-4y+10y ?
The extreme f(x,y)=3x^2+2y^2-4y+10y is Minimum(0,-3/2)