What is the extreme f(x)=18xy-x^3-9y^2 ?

The extreme f(x)=18xy-x^3-9y^2 is Saddle(0,0),Maximum(6,6)

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extreme f(x)=18xy-x^3-9y^2

extreme points

f (x) = 18 x y - x^{3} - 9 y^{2}

Saddle (0, 0), Maximum (6, 6)

Solution steps

Find the critical points:

(0, 0), (6, 6)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 18

D (x, y) = 108 x - 324

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(6, 6) :

Maximum

Saddle (0, 0), Maximum (6, 6)

x^{4} - 4 x^{3} + 16 x

f (x) = 15 x^{4} - 4 x^{3}, - 2 \leq x \leq 2

f (x) = x - \frac{1}{2} \cdot y^{2} - \frac{1}{3} x^{3} + y + 6

f (x) = x^{4} - 8 x^{3} + 18 x^{2}

f (x) = 4 x^{3} - 3 x^{2} - 6 x + 3, (- 1, 10)

What is the extreme f(x)=18xy-x^3-9y^2 ?
The extreme f(x)=18xy-x^3-9y^2 is Saddle(0,0),Maximum(6,6)