What is the extreme f(x,y)=2x^2-8x+y^2-8y+2 ?

The extreme f(x,y)=2x^2-8x+y^2-8y+2 is Minimum(2,4)

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extreme f(x,y)=2x^2-8x+y^2-8y+2

extreme points

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

Minimum (2, 4)

Solution steps

Find the critical points:

(2, 4)

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

D (x, y) = 8

Check the sign of at each critical point

Check critical point

(2, 4) :

Minimum

Minimum (2, 4)

\frac{1}{x - 2}

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

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

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

f (x) = x^{2}, - 1 \leq x \leq 0

What is the extreme f(x,y)=2x^2-8x+y^2-8y+2 ?
The extreme f(x,y)=2x^2-8x+y^2-8y+2 is Minimum(2,4)