MHB Mr.Ask's question at Yahoo Answers (Two variables, domain and range)

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The function f(x,y) = 1/√(16 - x² - y²) is defined only when 16 - x² - y² > 0, which corresponds to the domain being the open disk D((0,0),4). Therefore, the domain of f is the set of all points (x,y) such that x² + y² < 16. The range of the function varies from 1/4 to infinity, with 1/4 included and infinity excluded. Thus, the range is [1/4, +∞). This analysis provides a clear understanding of the function's domain and range.
Fernando Revilla
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Hello MrAsk,

There exists $f(x,y)=\dfrac{1}{\sqrt{16-x^2-y^2}}$ if and only if $16-x^2-y^2>0$ or equivalently if and only if $x^2+y^2<4^2$ (open disk $D((0,0),4)$). That is, $$\boxed{\:\mbox{Dom } (f)=D((0,0),4)\:}$$ On the other hand, for every $(x,y)$ such that $x^2+y^2=r^2$ with $0\leq r<4$ we have $f(x,y)=\dfrac{1}{\sqrt{16-r^2}}$ so, $f(x,y)$ varies from $1/4$ (included) to $+\infty$ (excluded). That is, $$\boxed{\:\mbox{Range } (f)=\left[1/4,+\infty\right)\:}$$
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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