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

The extreme 2x^2+y^2+xy-3x is Minimum(6/7 ,-3/7)

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

extreme points

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

Minimum (\frac{6}{7}, - \frac{3}{7})

Solution steps

Find the critical points:

(\frac{6}{7}, - \frac{3}{7})

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

D (x, y) = 7

Check the sign of at each critical point

Check critical point

(\frac{6}{7}, - \frac{3}{7}) :

Minimum

Minimum (\frac{6}{7}, - \frac{3}{7})

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

f (x) = x^{4} - 72 x^{2}

f (x) = - 3 x^{4} ln (3 x), (0, 4)

(2 x - 1)^{2}

What is the extreme 2x^2+y^2+xy-3x ?
The extreme 2x^2+y^2+xy-3x is Minimum(6/7 ,-3/7)