What is the extreme f(x,y)=7x-8y+2xy-x^2+y^1 ?

The extreme f(x,y)=7x-8y+2xy-x^2+y^1 is Saddle(7/2 ,0)

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extreme f(x,y)=7x-8y+2xy-x^2+y^1

extreme points

f (x, y) = 7 x - 8 y + 2 x y - x^{2} + y^{1}

Saddle (\frac{7}{2}, 0)

Solution steps

Find the critical points:

(\frac{7}{2}, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 0

D (x, y) = - 4

Check the sign of at each critical point

Check critical point

(\frac{7}{2}, 0) :

Saddle

Saddle (\frac{7}{2}, 0)

f (x) = \frac{6}{x}

f (x, y) = (x - y) (9 - x y)

f (x) = x + \frac{49}{x} + 9, 1 \leq x \leq 98

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

f (x) = x^{2} + \frac{200}{x}

What is the extreme f(x,y)=7x-8y+2xy-x^2+y^1 ?
The extreme f(x,y)=7x-8y+2xy-x^2+y^1 is Saddle(7/2 ,0)