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) = \frac{4 x}{x ^{2} + 9}

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

f (x) = x^{2} ln (x), \frac{1}{3} \leq x \leq \frac{5}{6}

f (x) = - x^{4} + 6 x^{2}

f (x) = \frac{2}{x} + 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)