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In a school, an auditorium was used for its cultural activities.  
The shape of the floor of the auditorium is rectangular with dimensions $x$ and $y$ $(x>y)$ and has fixed parameter $p$.

Based on the above information answer the following questions.

The value of $x$, for which area of floor of auditorium is maximum is:





Solution

$p = 2(x+y)$

So

$x+y=\frac{p}{2}$

Hence

$y=\frac{p}{2}-x$

Area of rectangle

$A=xy$

$A=x\left(\frac{p}{2}-x\right)$

$A=\frac{px}{2}-x^2$

To maximize area

$\frac{dA}{dx}=\frac{p}{2}-2x$

Set derivative equal to zero

$\frac{p}{2}-2x=0$

$2x=\frac{p}{2}$

$x=\frac{p}{4}$



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