What is the extreme f(xy)=x^2+2y^2-xy+14y ?

The extreme f(xy)=x^2+2y^2-xy+14y is Minimum(-2,-4)

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extreme f(xy)=x^2+2y^2-xy+14y

extreme points

f (x y) = x^{2} + 2 y^{2} - x y + 14 y

Minimum (- 2, - 4)

Solution steps

Find the critical points:

(- 2, - 4)

\frac{\partial ^{2} f}{\partial x ^{2}} = 4

D (x, y) = 7

Check the sign of at each critical point

Check critical point

(- 2, - 4) :

Minimum

Minimum (- 2, - 4)

\frac{x + 1}{x ^{2} - 7 x - 8}

f (x) = x^{3} - 12 x^{2} + 45 x + 4

f (x) = x^{3} - 12 x^{2} + 45 x + 6

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

f (x) = 2 x^{3} - 2 x^{2} - 2 x + 5

What is the extreme f(xy)=x^2+2y^2-xy+14y ?
The extreme f(xy)=x^2+2y^2-xy+14y is Minimum(-2,-4)