Maximum area of the rectangle that can be formed with the fixed perimeter ' \(p\) ' \(\mathrm{cm}\)
Maximum area of the rectangle that can be formed with the fixed perimeter ' \(p\) ' \(\mathrm{cm}\)
\(\frac{p^2}{8} \mathrm{~cm}^2\)
\(\frac{p^2}{16} \mathrm{~cm}^2\)
\(\frac{p^2}{64} \mathrm{~cm}^2\)
\(\frac{p^2}{32} \mathrm{~cm}^2\)
Solution
Let length of adjacent sides of rectangle is \(x \mathrm{~cm}\) and \(y \mathrm{~cm}\) so perimeter of rectangle is \(p=2(x+y) \mathrm{cm}\) and the area \(A=x y \mathrm{~cm}^2\)
\(\Rightarrow \quad A=x\left(\frac{p}{2}-x\right)\)
for maxima \(\frac{d A}{d x}=0 \Rightarrow \frac{p}{2}-2 x=0 \Rightarrow x=\frac{p}{4} \mathrm{~cm}\) and \(y=\frac{p}{4} \mathrm{~cm}\)
\(\therefore\) For given perimeter of rectangle \(p\), the maximum possible area \(A=\frac{p}{4} \times \frac{p}{4}=\frac{p^2}{16} \mathrm{~cm}^2\)
Hence, option (b) is correct.