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Find the maximum of $$\iint\limits_{D}~dx\,dy$$ as a function of $m$, $0<m<1$, where $D=\left\{(x,y): \frac{x^2}{m}+\frac{y^2}{1-m} \leq 1\right\}$.

Here $f(x,y)=\iint\limits_{D}~dx\,dy$. Now for maximum $\frac{\partial f}{\partial x}=0 \text{ and } \frac{\partial f}{\partial y}=0$. But here I am stuck. Please help.

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You are maximizing over all possible values of $m$. So, the variable is $m$, not $x$ or $y$. Solving $\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}=0$ is irrelevant. –  user1551 Jan 1 '13 at 10:02
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up vote 8 down vote accepted

$\textbf{Hint:}$ Note that $\iint\limits_{D}~dx dy$ is nothing but the area of the region $D$ enclosed by the ellipse, which is equal to $\pi\sqrt {m(1-m)}.$

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If $f(x,y)=\pi \sqrt{m(1-m)}$,then what is meant by "Maximum of $\pi \sqrt{m(1-m)}$ as a function of $m$." –  A.D Jan 1 '13 at 9:51
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@A.D: Note that thequestion says $\textit{Find the maximum of ... as a function of } m$, so you need to consider the function $f(m)=\pi\sqrt{m(1-m)}$ and find the value of $m$ which maximizes it. –  pritam Jan 1 '13 at 9:56
    
Thank you all. Now it is clear to me. –  A.D Jan 1 '13 at 9:58
    
. . . . and the value of $m$ that maximizes $\pi\sqrt{m(1-m)}$ is the same as the value of $m$ that maximizes $m(1-m)$. –  Michael Hardy Jan 1 '13 at 13:04
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