What is the extreme 3x^2+y^2-2xy+42-8y ?

The extreme 3x^2+y^2-2xy+42-8y is Minimum(2,6)

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extreme 3x^2+y^2-2xy+42-8y

extreme points

3 x^{2} + y^{2} - 2 x y + 42 - 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) = - \frac{56}{81 ( x + 1 ) ^{\frac{10}{3}}}

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

f (x) = \frac{- 1}{x - 7}

y = 2 x^{3} - 6 x + 3

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

What is the extreme 3x^2+y^2-2xy+42-8y ?
The extreme 3x^2+y^2-2xy+42-8y is Minimum(2,6)