What is the extreme f(x,y)=2x^2+3y^2+2xy+10x-20y ?

The extreme f(x,y)=2x^2+3y^2+2xy+10x-20y is Minimum(-5,5)

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extreme f(x,y)=2x^2+3y^2+2xy+10x-20y

extreme points

f (x, y) = 2 x^{2} + 3 y^{2} + 2 x y + 10 x - 20 y

Minimum (- 5, 5)

Solution steps

Find the critical points:

(- 5, 5)

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

D (x, y) = 20

Check the sign of at each critical point

Check critical point

(- 5, 5) :

Minimum

Minimum (- 5, 5)

f (x, y) = 48 x y - 32 x^{3} - 24 y^{2}

f (x, y) = x^{2} + x y + y^{2} - 34 y + 385

f (x, y) = 5 - 2 x + 4 y - x^{2} - 4 y^{2}

f (x) = x - 2 x

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

What is the extreme f(x,y)=2x^2+3y^2+2xy+10x-20y ?
The extreme f(x,y)=2x^2+3y^2+2xy+10x-20y is Minimum(-5,5)