What is the extreme f(x,y)=(9x^2+4x+5)(9y^2+6y+6) ?

The extreme f(x,y)=(9x^2+4x+5)(9y^2+6y+6) is Minimum(-2/9 ,-1/3)

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extreme f(x,y)=(9x^2+4x+5)(9y^2+6y+6)

extreme points

f (x, y) = (9 x^{2} + 4 x + 5) (9 y^{2} + 6 y + 6)

Minimum (- \frac{2}{9}, - \frac{1}{3})

Solution steps

Find the critical points:

(- \frac{2}{9}, - \frac{1}{3})

\frac{\partial ^{2} f}{\partial x ^{2}} = 18 (9 x^{2} + 4 x + 5)

D (x, y) = 5832 x^{2} - 78732 x^{2} y^{2} - 52488 x^{2} y + 2592 x - 34992 x y^{2} - 23328 x y + 9396 y^{2} + 6264 y + 9144

Check the sign of at each critical point

Check critical point

(- \frac{2}{9}, - \frac{1}{3}) :

Minimum

Minimum (- \frac{2}{9}, - \frac{1}{3})

f (x) = 6 x^{3} - x^{2} - 5 x + 7

f (x) = γ (x + 1)

f (x) = \frac{x}{x ^{2} - 64}

f (x) = \frac{x}{x ^{2} - x + 2}, - 2 \leq x \leq 1

- 4.9 t^{2} + 283 t + 353

What is the extreme f(x,y)=(9x^2+4x+5)(9y^2+6y+6) ?
The extreme f(x,y)=(9x^2+4x+5)(9y^2+6y+6) is Minimum(-2/9 ,-1/3)