What is the extreme f(x,y)=sqrt(107-4x^2-3y^2) ?

The extreme f(x,y)=sqrt(107-4x^2-3y^2) is Maximum(0,0)

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extreme f(x,y)=sqrt(107-4x^2-3y^2)

extreme points

f (x, y) = 107 - 4 x^{2} - 3 y^{2}

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{3 ( 107 - 4 x ^{2} )}{( 107 - 4 x ^{2} - 3 y ^{2} ) 107 - 4 x ^{2} - 3 y ^{2}}

D (x, y) = \frac{1284}{( 107 - 4 x ^{2} - 3 y ^{2} ) ^{2}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

f (x) = \frac{x ^{2} - x + 1}{x ^{2} - 2 x + 2}

f (x) = 3 + 3 x + x^{2}

\frac{x}{ln ( x )}, 2 \leq x \leq 10

f (x) = 10000 e^{0.17 (x - 1.62)^{2}}

f (x, y) = \frac{5 x}{1 + x ^{2} + y ^{2}}

What is the extreme f(x,y)=sqrt(107-4x^2-3y^2) ?
The extreme f(x,y)=sqrt(107-4x^2-3y^2) is Maximum(0,0)