What is the extreme f(x,y)=2x^2-8x+y^2-8y+6 ?

The extreme f(x,y)=2x^2-8x+y^2-8y+6 is Minimum(2,4)

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extreme f(x,y)=2x^2-8x+y^2-8y+6

extreme points

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

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)

f (x) = x e^{- 4 x}, 0 \leq x \leq 2

f (x) = - 1 - x + x^{2}

f (x) = - x^{2} + 4, - 2 \leq x \leq 3

x^{\frac{1}{5}} (x + 6)

f (x) = x^{2} e^{4 x}

What is the extreme f(x,y)=2x^2-8x+y^2-8y+6 ?
The extreme f(x,y)=2x^2-8x+y^2-8y+6 is Minimum(2,4)