What is the extreme f(x,y)=sqrt(49-(x-9)^2-y^2) ?

The extreme f(x,y)=sqrt(49-(x-9)^2-y^2) is Maximum(9,0)

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extreme f(x,y)=sqrt(49-(x-9)^2-y^2)

extreme points

f (x, y) = 49 - (x - 9)^{2} - y^{2}

Maximum (9, 0)

Solution steps

Find the critical points:

(9, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{- x ^{2} + 18 x - 32}{( - x ^{2} + 18 x - y ^{2} - 32 ) - x ^{2} + 18 x - y ^{2} - 32}

D (x, y) = \frac{49}{( x ^{2} - 18 x + y ^{2} + 32 ) ^{2}}

Check the sign of at each critical point

Check critical point

(9, 0) :

Maximum

Maximum (9, 0)

f (x, y) = x^{2} y^{2} + 2 x y

f (x) = 4 x^{2} - 5 x - 2

f (x, y) = x y - 2 x - 4 y

f (x) = 4 x^{2} - 5 x + 1

What is the extreme f(x,y)=sqrt(49-(x-9)^2-y^2) ?
The extreme f(x,y)=sqrt(49-(x-9)^2-y^2) is Maximum(9,0)