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

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

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

extreme points

f (x) = 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)

x^{3} - 3 x^{2} + 3 x

f (x, y) = y - x + 2

f (x) = x^{3} - 3 x^{2} - 9 x + 8

f (x) = x^{3} - 3 x^{2} - 9 x + 9

f (x) = x^{3} - 12 x, - 3 \leq x \leq 3

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