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, y) = 4 - x^{2} - 4 y^{2}

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

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

y = x^{3} - 3 x + 1

f (x) = - 2 x^{4} + 8 x^{3} - 1

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)