AP EAMCET · Maths · Straight Lines
Maximum area of the rectangle that can be formed with the fixed perimeter ' \(p\) ' \(\mathrm{cm}\)
- A \(\frac{p^2}{8} \mathrm{~cm}^2\)
- B \(\frac{p^2}{16} \mathrm{~cm}^2\)
- C \(\frac{p^2}{64} \mathrm{~cm}^2\)
- D \(\frac{p^2}{32} \mathrm{~cm}^2\)
Answer & Solution
Correct Answer
(B) \(\frac{p^2}{16} \mathrm{~cm}^2\)
Step-by-step Solution
Detailed explanation
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…
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