What is the extreme f(x,y)=x^2+2xy^2+(2y)/(3x) ?

The extreme f(x,y)=x^2+2xy^2+(2y)/(3x) is Saddle(-1/(2^{2/5)*3^{1/5}},-1/(\sqrt[5]{54)})

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extreme f(x,y)=x^2+2xy^2+(2y)/(3x)

extreme points

f (x, y) = x^{2} + 2 x y^{2} + \frac{2 y}{3 x}

Saddle (- \frac{1}{2 ^{\frac{2}{5}} \cdot 3 ^{\frac{1}{5}}}, - \frac{1}{5 54})

Solution steps

Find the critical points:

(- \frac{1}{2 ^{\frac{2}{5}} \cdot 3 ^{\frac{1}{5}}}, - \frac{1}{5 54})

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

D (x, y) = 8 x + \frac{32 y}{3 x ^{2}} - \frac{4}{9 x ^{4}} - 16 y^{2}

Check the sign of at each critical point

Check critical point

(- \frac{1}{2 ^{\frac{2}{5}} \cdot 3 ^{\frac{1}{5}}}, - \frac{1}{5 54}) :

Saddle

Saddle (- \frac{1}{2 ^{\frac{2}{5}} \cdot 3 ^{\frac{1}{5}}}, - \frac{1}{5 54})

P (x, y) = 30 x^{2} + 2 x y - 4 y^{2} + 47 x - 12 y + 7

f (x, y) = 2 x^{2} + 3 x y + 4 y^{2} - 5 x + 2 y

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

f (x) = x^{\frac{3}{5}} (4 - x)

f (x) = x y + \frac{50}{x} + \frac{20}{y}

What is the extreme f(x,y)=x^2+2xy^2+(2y)/(3x) ?
The extreme f(x,y)=x^2+2xy^2+(2y)/(3x) is Saddle(-1/(2^{2/5)*3^{1/5}},-1/(\sqrt[5]{54)})