What is the extreme f(x,y)=ln(3-4x^2-5y^2) ?

The extreme f(x,y)=ln(3-4x^2-5y^2) is Maximum(0,0)

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extreme f(x,y)=ln(3-4x^2-5y^2)

extreme points

f (x, y) = ln (3 - 4 x^{2} - 5 y^{2})

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{10 ( 3 - 4 x ^{2} + 5 y ^{2} )}{( 3 - 4 x ^{2} - 5 y ^{2} ) ^{2}}

D (x, y) = - \frac{80 ( 4 x ^{2} + 5 y ^{2} + 3 )}{( 4 x ^{2} + 5 y ^{2} - 3 ) ^{3}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

y = x^{2} + 9 x - 2

f (x) = 5 cos^{2} (x), (0, π)

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

f (x, y) = 50 + 2 x + 3 y - 3 x^{2} - y^{2} - 0.5 x y

U (x, y) = - x^{2} - 3 y^{2} + 2 x y + 100 x

What is the extreme f(x,y)=ln(3-4x^2-5y^2) ?
The extreme f(x,y)=ln(3-4x^2-5y^2) is Maximum(0,0)