What is the extreme f(x,y)=2x^2+3xy+4y^2+7x+11y ?

The extreme f(x,y)=2x^2+3xy+4y^2+7x+11y is Minimum(-1,-1)

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extreme f(x,y)=2x^2+3xy+4y^2+7x+11y

extreme points

f (x, y) = 2 x^{2} + 3 x y + 4 y^{2} + 7 x + 11 y

Minimum (- 1, - 1)

Solution steps

Find the critical points:

(- 1, - 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = 8

D (x, y) = 23

Check the sign of at each critical point

Check critical point

(- 1, - 1) :

Minimum

Minimum (- 1, - 1)

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

2 x^{3} + 6 x y^{2} - 3 y^{3} - 150 x

f (x) = x^{3} - 2 x^{2} - 15 x + 10, - 2 \leq x \leq 0

f (x, y) = 2 x^{4} - 5 x^{2} y^{3} - 3 x y^{4}

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

What is the extreme f(x,y)=2x^2+3xy+4y^2+7x+11y ?
The extreme f(x,y)=2x^2+3xy+4y^2+7x+11y is Minimum(-1,-1)