What is the extreme f(x,y)=x^2+2y^2+xy-5x-6y ?

The extreme f(x,y)=x^2+2y^2+xy-5x-6y is Minimum(2,1)

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extreme f(x,y)=x^2+2y^2+xy-5x-6y

extreme points

f (x, y) = x^{2} + 2 y^{2} + x y - 5 x - 6 y

Minimum (2, 1)

Solution steps

Find the critical points:

(2, 1)

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

D (x, y) = 7

Check the sign of at each critical point

Check critical point

(2, 1) :

Minimum

Minimum (2, 1)

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

f (x) = \frac{x ^{2} - 5}{x + 3}

f (x, y) = 2 x^{3} + y^{2} - 6 x y - 23

f (x) = \frac{x}{x - 4}, 6 \leq x \leq 10

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

What is the extreme f(x,y)=x^2+2y^2+xy-5x-6y ?
The extreme f(x,y)=x^2+2y^2+xy-5x-6y is Minimum(2,1)