What is the extreme f(x,y)= 9/(sqrt(6-4x^2-9y^2)) ?

The extreme f(x,y)= 9/(sqrt(6-4x^2-9y^2)) is Minimum(0,0)

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extreme f(x,y)= 9/(sqrt(6-4x^2-9y^2))

extreme points

f (x, y) = \frac{9}{6 - 4 x ^{2} - 9 y ^{2}}

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{81 ( ( 6 - 4 x ^{2} - 9 y ^{2} ) ^{\frac{3}{2}} + 27 y ^{2} - 9 y ^{2} + 6 - 4 x ^{2} )}{( 6 - 4 x ^{2} - 9 y ^{2} ) ^{3}}

D (x, y) = \frac{- 186624 x ^{6} - 1259712 x ^{4} y ^{2} + 839808 x ^{4} - 2834352 x ^{2} y ^{4} + 3779136 x ^{2} y ^{2} - 1259712 x ^{2} + 34992 x ^{2} ( - 4 x ^{2} - 9 y ^{2} + 6 ) ^{\frac{3}{2}} - 4 x ^{2} - 9 y ^{2} + 6 + 78732 y ^{2} ( - 4 x ^{2} - 9 y ^{2} + 6 ) ^{\frac{3}{2}} - 4 x ^{2} - 9 y ^{2} + 6 - 2125764 y ^{6} + 4251528 y ^{4} - 2834352 y ^{2} + 629856}{( 6 - 4 x ^{2} - 9 y ^{2} ) ^{6}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

f (x, y) = x^{3} + y^{3} - 33 x y

f (x) = 5 x^{5} - 100

\frac{2 - x}{e ^{x}}

f (x) = 13 x^{3} + 138 x^{2} - 103

f (x) = x^{3} + 6 x^{2} - 36 x

What is the extreme f(x,y)= 9/(sqrt(6-4x^2-9y^2)) ?
The extreme f(x,y)= 9/(sqrt(6-4x^2-9y^2)) is Minimum(0,0)