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

The extreme f(x,y)=x^2+y^2-2x-4y is Minimum(1,2)

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

extreme points

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

Minimum (1, 2)

Solution steps

Find the critical points:

(1, 2)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(1, 2) :

Minimum

Minimum (1, 2)

f (x) = - 7 x^{2} - 7 x y - 6 y^{2} - 35 x - 26 y + 9

3 (x - 4)^{\frac{2}{3}} + 6

y = x^{3} + 3 x^{2}

f (x) = 4 x^{\frac{3}{5}}

f (x) = x - (\frac{x}{y}) \cdot y

What is the extreme f(x,y)=x^2+y^2-2x-4y ?
The extreme f(x,y)=x^2+y^2-2x-4y is Minimum(1,2)