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

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

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

extreme points

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

Minimum (2, 6)

Solution steps

Find the critical points:

(2, 6)

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

D (x, y) = 8

Check the sign of at each critical point

Check critical point

(2, 6) :

Minimum

Minimum (2, 6)

f (x) = - 5 x + 3 ln (4 x)

f (x, y) = x y + 4 x

g (t) = 3 [u (t - 6) - u (t - 7)]

f (x) = \frac{3 x}{x ^{2} + 1}

f (x) = \frac{x ^{3} + 8}{x}

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