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

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

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

extreme points

f (x) = 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)

f (x, y) = 6 x + 3 y - 7

- 2 x^{2} + 3 x - 1

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

f (x, y) = 1 - x y

f (x) = \frac{1}{7} x^{7} - x

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