What is the extreme f(x,y)=2x^2+3y^2-12x-2y+1 ?

The extreme f(x,y)=2x^2+3y^2-12x-2y+1 is Minimum(3, 1/3)

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extreme f(x,y)=2x^2+3y^2-12x-2y+1

extreme points

f (x, y) = 2 x^{2} + 3 y^{2} - 12 x - 2 y + 1

Minimum (3, \frac{1}{3})

Solution steps

Find the critical points:

(3, \frac{1}{3})

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

D (x, y) = 24

Check the sign of at each critical point

Check critical point

(3, \frac{1}{3}) :

Minimum

Minimum (3, \frac{1}{3})

x^{2} (x - 1) (2 x + 1) + y^{2}

f (x, y) = \frac{x y}{x ^{2} + y ^{2} - 25}

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

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

f (x) = - x^{2} + 8

What is the extreme f(x,y)=2x^2+3y^2-12x-2y+1 ?
The extreme f(x,y)=2x^2+3y^2-12x-2y+1 is Minimum(3, 1/3)