What is the extreme f(x,y)=4-x^2-y^2+xy+x ?

The extreme f(x,y)=4-x^2-y^2+xy+x is Maximum(2/3 , 1/3)

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extreme f(x,y)=4-x^2-y^2+xy+x

extreme points

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

Maximum (\frac{2}{3}, \frac{1}{3})

Solution steps

Find the critical points:

(\frac{2}{3}, \frac{1}{3})

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

D (x, y) = 3

Check the sign of at each critical point

Check critical point

(\frac{2}{3}, \frac{1}{3}) :

Maximum

Maximum (\frac{2}{3}, \frac{1}{3})

f (x) = \frac{x ^{2}}{x ^{2} + 3}, - 1 \leq x \leq 1

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

f (x) = \frac{x ^{2}}{x ^{2} + 1}, - 2 \leq x \leq 2

f (x) = x 1 - 5 x

f (x, y) = 9 x^{2} + 3 x y + 9 y^{2} + 7 x y^{2} + 3 x^{2} y

What is the extreme f(x,y)=4-x^2-y^2+xy+x ?
The extreme f(x,y)=4-x^2-y^2+xy+x is Maximum(2/3 , 1/3)