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

The extreme f(x,y)=(2y-x)/((x-1)^2+(y-2)^2) is Minimum(11/5 ,-2/5)

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extreme f(x,y)=(2y-x)/((x-1)^2+(y-2)^2)

extreme points

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

Minimum (\frac{11}{5}, - \frac{2}{5})

Solution steps

Find the critical points:

(\frac{11}{5}, - \frac{2}{5})

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{2 ( x ^{3} + 6 x ^{2} - 6 x ^{2} y + 24 x y - 3 x y ^{2} - 27 x + 2 y ^{3} + 40 - 30 y )}{( x ^{2} - 2 x + y ^{2} + 5 - 4 y ) ^{3}}

D (x, y) = \frac{4 ( - 2 y ^{3} + 3 x y ^{2} + 6 y ^{2} + 6 x ^{2} y + 6 y - 24 x y + 15 x - x ^{3} - 10 ) ( x ^{3} + 6 x ^{2} - 6 x ^{2} y + 24 x y - 3 x y ^{2} - 27 x + 2 y ^{3} + 40 - 30 y ) - 4 ( - 2 x ^{3} + 12 x ^{2} - 3 x ^{2} y + 6 x y ^{2} - 6 x y - 18 x + y ^{3} + 21 y - 12 y ^{2} ) ^{2}}{( x ^{2} + y ^{2} + 5 - 2 x - 4 y ) ^{6}}

Check the sign of at each critical point

Check critical point

(\frac{11}{5}, - \frac{2}{5}) :

Minimum

Minimum (\frac{11}{5}, - \frac{2}{5})

f (x, y) = x - 1 y - 1

x (x + 2)^{3}

f (x, y) = 8 x^{3} + 2 x y - 3 x^{2} + y^{2} + 1

f (x) = 3 x + 4 y

\frac{1 - x ^{2}}{x}

What is the extreme f(x,y)=(2y-x)/((x-1)^2+(y-2)^2) ?
The extreme f(x,y)=(2y-x)/((x-1)^2+(y-2)^2) is Minimum(11/5 ,-2/5)