What is the extreme f(x,y)=2x^2-4x+7y^2+8 ?

The extreme f(x,y)=2x^2-4x+7y^2+8 is Minimum(1,0)

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extreme f(x,y)=2x^2-4x+7y^2+8

extreme points

f (x, y) = 2 x^{2} - 4 x + 7 y^{2} + 8

Minimum (1, 0)

Solution steps

Find the critical points:

(1, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 14

D (x, y) = 56

Check the sign of at each critical point

Check critical point

(1, 0) :

Minimum

Minimum (1, 0)

f (x) = 1 + \frac{1}{x} + \frac{7}{x ^{2}} + \frac{1}{x ^{3}}

f (x) = x 49 - x^{2}, - 7 \leq x \leq 7

f (x) = - 0.1 t^{2} + 1.4 t + 98.5

f (x) = 3 x^{2} - 8 x^{3} + 6 x^{2} + 3

f (x) = - e^{(a (4 - x))} + 1

What is the extreme f(x,y)=2x^2-4x+7y^2+8 ?
The extreme f(x,y)=2x^2-4x+7y^2+8 is Minimum(1,0)