What is the extreme f(x,y)=5x^2+3xy+y^2 ?

The extreme f(x,y)=5x^2+3xy+y^2 is Minimum(0,0)

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extreme f(x,y)=5x^2+3xy+y^2

extreme points

f (x, y) = 5 x^{2} + 3 x y + y^{2}

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = 11

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

f (x, y) = x + y - x y

\frac{- 3 x + 48}{( x - 120 ) ^{9}}

f (x, y) = x^{2} + x y + y^{2} + x + a y

e^{x}

f (x) = x^{2} + y^{2} + 5

What is the extreme f(x,y)=5x^2+3xy+y^2 ?
The extreme f(x,y)=5x^2+3xy+y^2 is Minimum(0,0)