What is the extreme f(x,y)=-x^3+8xy-6y^2+12 ?

The extreme f(x,y)=-x^3+8xy-6y^2+12 is Saddle(0,0),Maximum(16/9 , 32/27)

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extreme f(x,y)=-x^3+8xy-6y^2+12

extreme points

f (x, y) = - x^{3} + 8 x y - 6 y^{2} + 12

Saddle (0, 0), Maximum (\frac{16}{9}, \frac{32}{27})

Solution steps

Find the critical points:

(0, 0), (\frac{16}{9}, \frac{32}{27})

\frac{\partial ^{2} f}{\partial x ^{2}} = - 12

D (x, y) = 72 x - 64

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(\frac{16}{9}, \frac{32}{27}) :

Maximum

Saddle (0, 0), Maximum (\frac{16}{9}, \frac{32}{27})

(x - 1)^{3} + 2

f (x) = x^{5} \cdot ln (x)

f (x) = ((42 - x) x - (2 x + 80))

f (x) = (x - 8) ln (x - 8)

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

What is the extreme f(x,y)=-x^3+8xy-6y^2+12 ?
The extreme f(x,y)=-x^3+8xy-6y^2+12 is Saddle(0,0),Maximum(16/9 , 32/27)