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

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

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

extreme points

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

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = 224

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

x x + 7

f (y) = x y

f (x) = - x^{2} + 120 x - 3200

2 x^{3} - 3 x^{2} - 36 x + 54

2 x^{2} - x y + y^{2} + 7 x

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