What is the extreme f(x,y)=3-sqrt(25-(x-2)^2-4(y+1)^2) ?

The extreme f(x,y)=3-sqrt(25-(x-2)^2-4(y+1)^2) is Minimum(2,-1)

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extreme f(x,y)=3-sqrt(25-(x-2)^2-4(y+1)^2)

extreme points

f (x, y) = 3 - 25 - (x - 2)^{2} - 4 (y + 1)^{2}

Minimum (2, - 1)

Solution steps

Find the critical points:

(2, - 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{4 ( - x ^{2} + 4 x + 21 )}{( - x ^{2} + 4 x + 17 - 4 y ^{2} - 8 y ) - 4 y ^{2} - 8 y + 4 x + 17 - x ^{2}}

D (x, y) = \frac{100}{( x ^{2} - 4 x + 4 y ^{2} + 8 y - 17 ) ^{2}}

Check the sign of at each critical point

Check critical point

(2, - 1) :

Minimum

Minimum (2, - 1)

f (x) = 4 x^{2} - \frac{1}{x}

f (x) = x e^{(\frac{x}{2})}, - 3 \leq x \leq 1

f (x) = \frac{x ^{2}}{x ^{2} - 64}

f (x) = - x - (\frac{64}{x})

f (x) = - 0.3 x^{2} + 42 x + 200

What is the extreme f(x,y)=3-sqrt(25-(x-2)^2-4(y+1)^2) ?
The extreme f(x,y)=3-sqrt(25-(x-2)^2-4(y+1)^2) is Minimum(2,-1)