What is the extreme f(x,y)=x^3+9y^4-20x^2y+2 ?

The extreme f(x,y)=x^3+9y^4-20x^2y+2 is Minimum(320000/243 ,(320000^2)/(1036800000))

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extreme f(x,y)=x^3+9y^4-20x^2y+2

extreme points

f (x, y) = x^{3} + 9 y^{4} - 20 x^{2} y + 2

Minimum (\frac{320000}{243}, \frac{32000 0 ^{2}}{1036800000})

Solution steps

Find the critical points:

(\frac{320000}{243}, \frac{32000 0 ^{2}}{1036800000})

\frac{\partial ^{2} f}{\partial x ^{2}} = 108 y^{2}

D (x, y) = 108 y^{2} (6 x - 40 y) - 1600 x^{2}

Check the sign of at each critical point

Check critical point

(\frac{320000}{243}, \frac{32000 0 ^{2}}{1036800000}) :

Minimum

Minimum (\frac{320000}{243}, \frac{32000 0 ^{2}}{1036800000})

2 x^{3} + 3 x^{2} + 4

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

f (x) = \frac{2 - x}{5 + x}

f (x) = x^{(\frac{2}{5})} (\frac{7}{2 + x})

Q (x, y) = 5 x + 3 x y

What is the extreme f(x,y)=x^3+9y^4-20x^2y+2 ?
The extreme f(x,y)=x^3+9y^4-20x^2y+2 is Minimum(320000/243 ,(320000^2)/(1036800000))