What is the extreme f(x,y)=2x^2+3y^2+2xy-6x-8y+11 ?

The extreme f(x,y)=2x^2+3y^2+2xy-6x-8y+11 is Minimum(1,1)

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extreme f(x,y)=2x^2+3y^2+2xy-6x-8y+11

extreme points

f (x, y) = 2 x^{2} + 3 y^{2} + 2 x y - 6 x - 8 y + 11

Minimum (1, 1)

Solution steps

Find the critical points:

(1, 1)

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

D (x, y) = 20

Check the sign of at each critical point

Check critical point

(1, 1) :

Minimum

Minimum (1, 1)

f (x, y) = x^{2} - 2 x y + \frac{1}{3} y^{3} - 4 y

f (x) = 4 x^{5} - 25 x^{4} + 40 x^{3} - 4 [- 1.4]

f (x) = - \frac{1}{2} x^{2} + 15 x - 51

f (x) = (x^{2} - \frac{1}{2}) e^{- 2 x}

f (x) = x^{2} + y^{2} - 16 = 0

What is the extreme f(x,y)=2x^2+3y^2+2xy-6x-8y+11 ?
The extreme f(x,y)=2x^2+3y^2+2xy-6x-8y+11 is Minimum(1,1)