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

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

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

extreme points

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

Minimum (1, 1)

Solution steps

Find the critical points:

(1, 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = 4

D (x, y) = 8

Check the sign of at each critical point

Check critical point

(1, 1) :

Minimum

Minimum (1, 1)

f (x) = 0.00397 t^{2} - 0.593 t + 33.718

f (x) = f (x) = e x - x, - 4 \leq x \leq 2

f (x) = 4 x^{2} - 48 x + 190

f (x, y) = e^{9 x^{2} + 6 y^{2} + 1}

f (x) = (\frac{3 - cos ( θ )}{sin ( θ )})

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