What is the extreme f(x)=16xy-x^3-8y^2 ?

The extreme f(x)=16xy-x^3-8y^2 is Saddle(0,0),Maximum(16/3 , 16/3)

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extreme f(x)=16xy-x^3-8y^2

extreme points

f (x) = 16 x y - x^{3} - 8 y^{2}

Saddle (0, 0), Maximum (\frac{16}{3}, \frac{16}{3})

Solution steps

Find the critical points:

(0, 0), (\frac{16}{3}, \frac{16}{3})

\frac{\partial ^{2} f}{\partial x ^{2}} = - 16

D (x, y) = 96 x - 256

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(\frac{16}{3}, \frac{16}{3}) :

Maximum

Saddle (0, 0), Maximum (\frac{16}{3}, \frac{16}{3})

f (x) = x + \frac{9}{x} + 3

f (x) = - x^{4} + 8 x^{2} + 10

f (x) = (x^{\frac{5}{2} - 1} \cdot e^{- \frac{x}{2}})

f (x, y) = 36 - 4 x^{2} - 9 y^{2}

f (x, y) = 242 y^{2} + x^{2} - x^{2} y

What is the extreme f(x)=16xy-x^3-8y^2 ?
The extreme f(x)=16xy-x^3-8y^2 is Saddle(0,0),Maximum(16/3 , 16/3)