What is the extreme f(x,y)=2x^2+3y^2-4x-6 ?

The extreme f(x,y)=2x^2+3y^2-4x-6 is Minimum(1,0)

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extreme f(x,y)=2x^2+3y^2-4x-6

extreme points

f (x, y) = 2 x^{2} + 3 y^{2} - 4 x - 6

Minimum (1, 0)

Solution steps

Find the critical points:

(1, 0)

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

D (x, y) = 24

Check the sign of at each critical point

Check critical point

(1, 0) :

Minimum

Minimum (1, 0)

f (x, y) = 400 - 100 x^{2} - 16 y^{2}

f (x, y) = x^{3} + y^{3} + 3 x y^{2} - 18 (x + y)

x, f = \frac{20000}{3 + x ^{2} + y ^{2}}

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

y = - 4 e^{- 4 t} - 5 e^{- 5 t}

What is the extreme f(x,y)=2x^2+3y^2-4x-6 ?
The extreme f(x,y)=2x^2+3y^2-4x-6 is Minimum(1,0)