What is the extreme f(x,y)=2x^2+xy+4y^2-2x+10y ?

The extreme f(x,y)=2x^2+xy+4y^2-2x+10y is Minimum(26/31 ,-42/31)

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extreme f(x,y)=2x^2+xy+4y^2-2x+10y

extreme points

f (x, y) = 2 x^{2} + x y + 4 y^{2} - 2 x + 10 y

Minimum (\frac{26}{31}, - \frac{42}{31})

Solution steps

Find the critical points:

(\frac{26}{31}, - \frac{42}{31})

\frac{\partial ^{2} f}{\partial x ^{2}} = 8

D (x, y) = 31

Check the sign of at each critical point

Check critical point

(\frac{26}{31}, - \frac{42}{31}) :

Minimum

Minimum (\frac{26}{31}, - \frac{42}{31})

f (x) = x^{3} - 6 x^{2} - 9 x - 2

f (x) = x \cdot 500 - x

f (x) = - 8 x^{2} + x^{4}

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

f (t) = 108 t - t^{3}, - 7 \leq t < \infty

What is the extreme f(x,y)=2x^2+xy+4y^2-2x+10y ?
The extreme f(x,y)=2x^2+xy+4y^2-2x+10y is Minimum(26/31 ,-42/31)