What is the extreme 3x^2-2xy+y^2-8y ?

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

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extreme 3x^2-2xy+y^2-8y

extreme points

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) = 32 e^{4 x} x - 28 e^{4 x}

f (x) = 12 x - 9 x^{2} + 2 x^{3}

f (x, y) = 5 x^{2} - 4 x y + 4 y^{2} + 12 x + 25

f (x) = 2 x^{3} - \frac{12 x ^{2}}{2} - 2 [- 1.3]

\frac{27 x ^{2}}{( 2 - x ) ^{3}}

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