What is the extreme f(x,y)=(e^{2-xy})/(x+y^2) ?

The extreme f(x,y)=(e^{2-xy})/(x+y^2) is Saddle(\sqrt[3]{3}-1/(3^{2/3)},-1/(\sqrt[3]{3)})

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extreme f(x,y)=(e^{2-xy})/(x+y^2)

extreme points

f (x, y) = \frac{e ^{2 - x y}}{x + y ^{2}}

Saddle (33 - \frac{1}{3 ^{\frac{2}{3}}}, - \frac{1}{3 3})

Solution steps

Find the critical points:

(33 - \frac{1}{3 ^{\frac{2}{3}}}, - \frac{1}{3 3})

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{e ^{- x y + 2} ( - x ^{4} - 2 x ^{3} y ^{2} - x ^{2} y ^{4} - 4 x ^{2} y + 2 x - 4 x y ^{3} - 6 y ^{2} )}{( x + y ^{2} ) ^{3}}

D (x, y) = \frac{- e ^{- 2 x y + 4} ( y ^{2} ( x + y ^{2} ) ^{2} + 2 ( y ( x + y ^{2} ) + 1 ) ) ( - x ^{4} - 2 x ^{3} y ^{2} - x ^{2} y ^{4} - 4 x ^{2} y + 2 x - 4 x y ^{3} - 6 y ^{2} ) - e ^{2 (- x y + 2)} y ^{2} ( x ^{3} + 2 x ^{2} y ^{2} + x y ^{4} + x y + y ^{3} + 4 ) ^{2}}{( y ^{2} + x ) ^{6}}

Check the sign of at each critical point

Check critical point

(33 - \frac{1}{3 ^{\frac{2}{3}}}, - \frac{1}{3 3}) :

Saddle

Saddle (33 - \frac{1}{3 ^{\frac{2}{3}}}, - \frac{1}{3 3})

f (x) = \frac{- 7}{x ^{2} - 9}

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

f (x, y) = e^{y} - y e^{x}

f (x) = 2 x + cos (2 x), 0 < x < π

f (x, y) = x^{2} - 3 x y + 5 x - 2 y + 6 y^{2} + 8

What is the extreme f(x,y)=(e^{2-xy})/(x+y^2) ?
The extreme f(x,y)=(e^{2-xy})/(x+y^2) is Saddle(\sqrt[3]{3}-1/(3^{2/3)},-1/(\sqrt[3]{3)})