What is the extreme f(x,y)=16x\sqrt[3]{y}-2x^2-4x-16y+2 ?

The extreme f(x,y)=16x\sqrt[3]{y}-2x^2-4x-16y+2 is Saddle(1/3 , 1/27),Maximum(3,1)

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extreme f(x,y)=16x\sqrt[3]{y}-2x^2-4x-16y+2

extreme

Saddle (\frac{1}{3}, \frac{1}{27}), Maximum (3, 1)

Solution steps

Find the critical points:

(\frac{1}{3}, \frac{1}{27}), (3, 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{32 x}{9 y ^{\frac{5}{3}}}

D (x, y) = \frac{128 x}{9 y ^{\frac{5}{3}}} - \frac{256}{9 y ^{\frac{4}{3}}}

Check the sign of at each critical point

Check critical point

(\frac{1}{3}, \frac{1}{27}) :

Saddle

Check critical point

(3, 1) :

Maximum

Saddle (\frac{1}{3}, \frac{1}{27}), Maximum (3, 1)

What is the extreme f(x,y)=16x\sqrt[3]{y}-2x^2-4x-16y+2 ?
The extreme f(x,y)=16x\sqrt[3]{y}-2x^2-4x-16y+2 is Saddle(1/3 , 1/27),Maximum(3,1)