What is the extreme f(x)=7x-8y+2xy-x^2+y^3 ?

The extreme f(x)=7x-8y+2xy-x^2+y^3 is Saddle(23/6 , 1/3),Maximum(5/2 ,-1)

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extreme f(x)=7x-8y+2xy-x^2+y^3

extreme points

f (x) = 7 x - 8 y + 2 x y - x^{2} + y^{3}

Saddle (\frac{23}{6}, \frac{1}{3}), Maximum (\frac{5}{2}, - 1)

Solution steps

Find the critical points:

(\frac{23}{6}, \frac{1}{3}), (\frac{5}{2}, - 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = 6 y

D (x, y) = - 12 y - 4

Check the sign of at each critical point

Check critical point

(\frac{23}{6}, \frac{1}{3}) :

Saddle

Check critical point

(\frac{5}{2}, - 1) :

Maximum

Saddle (\frac{23}{6}, \frac{1}{3}), Maximum (\frac{5}{2}, - 1)

What is the extreme f(x)=7x-8y+2xy-x^2+y^3 ?
The extreme f(x)=7x-8y+2xy-x^2+y^3 is Saddle(23/6 , 1/3),Maximum(5/2 ,-1)