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If from a wire of length 36 meter a rectangle of greatest area is made, then its two adjacent sides in meter are
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9, 9
$\begin{aligned} & \text { Given } 2(a+b)=36, \quad a+b=18 \\ & \text { Area of rectangle }=a b=a(18-a) \\ & A=18 a-a^2, \therefore \frac{d A}{d a}=18-2 a \\ & \text { Put } \frac{d A}{d a}=0 \text { i.e., } 18-2 a=0 \Rightarrow a=9 ; b=9\end{aligned}$
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