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

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

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

extreme points

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

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{8 ( 4 + 7 x ^{2} )}{( 4 + 7 x ^{2} + 8 y ^{2} ) 4 + 7 x ^{2} + 8 y ^{2}}

D (x, y) = \frac{224}{( 4 + 7 x ^{2} + 8 y ^{2} ) ^{2}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

f (x) = - x^{4} + 4 x^{2} - 2

f (x) = x^{3} + y^{3} - 9 x y + 27

- 45 x^{2} - 27 x + 2

y = x^{3} - 9 x^{2} - 48 x

f (x, y) = 100 (y - x^{2})^{2} + (1 - x)^{2}

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