Question

Maximum area of the rectangle that can be formed with the fixed perimeter ' $p^{\prime }cm$_________

• A. $\frac{p^2}{8}c{m}^2$
• B. $\frac{p^2}{16}c{m}^2$
• C. $\frac{p^2}{64}c{m}^2$
• D. $\frac{p^2}{32}c{m}^2$

Answer 1

Answer: (B)

Let length of adjacent sides of rectangle is $xcm$ and $ycm$ so perimeter of rectangle is $p=2(x+y)cm$ and the area $A=xyc{m}^2$
$\Rightarrow A=x\left(\frac{p}{2}-x\right)$
for maxima
$\frac{dA}{dx}=0\Rightarrow \frac{p}{2}-2x=0$
$\Rightarrow x=\frac{p}{4}cm$
and $y=\frac{p}{4}cm$
$\therefore$ For given perimeter of rectangle $p$, the maximum possible area
$A=\frac{p}{4}\times \frac{p}{4}=\frac{p^2}{16}c{m}^2$