What is the extreme f(x,y)=2x^2+4xy+6y^2-8x-10 ?

The extreme f(x,y)=2x^2+4xy+6y^2-8x-10 is Minimum(3,-1)

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extreme f(x,y)=2x^2+4xy+6y^2-8x-10

extreme points

f (x, y) = 2 x^{2} + 4 x y + 6 y^{2} - 8 x - 10

Minimum (3, - 1)

Solution steps

Find the critical points:

(3, - 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = 12

D (x, y) = 32

Check the sign of at each critical point

Check critical point

(3, - 1) :

Minimum

Minimum (3, - 1)

f (x, y) = x e^{- 2 x^{2} - 2 y^{2}}

4 x y

f (x) = \frac{- 2 x ^{2} + 5 x - 1}{2 x - 1}

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

f (x) = x^{2} (x + 1)^{3}

What is the extreme f(x,y)=2x^2+4xy+6y^2-8x-10 ?
The extreme f(x,y)=2x^2+4xy+6y^2-8x-10 is Minimum(3,-1)