What is the extreme f(x,y)=ln(1+4x^2+6y^2) ?

The extreme f(x,y)=ln(1+4x^2+6y^2) is Minimum(0,0)

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extreme f(x,y)=ln(1+4x^2+6y^2)

extreme points

f (x, y) = ln (1 + 4 x^{2} + 6 y^{2})

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{12 ( 1 + 4 x ^{2} - 6 y ^{2} )}{( 1 + 4 x ^{2} + 6 y ^{2} ) ^{2}}

D (x, y) = \frac{- 1536 x ^{4} - 4608 x ^{2} y ^{2} - 3456 y ^{4} + 96}{( 1 + 4 x ^{2} + 6 y ^{2} ) ^{4}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

f (x) = x^{2} - 6 x y + 2 y^{2} + 10 x + 2 y - 5

25 x

f (x) = x - \frac{250}{x}

f (x) = - 25 x^{2} - 25 y^{2} + 200 x + 200 y + 200

f (x) = \frac{( x + 8 )}{x ^{2}}

What is the extreme f(x,y)=ln(1+4x^2+6y^2) ?
The extreme f(x,y)=ln(1+4x^2+6y^2) is Minimum(0,0)