The area (in sq. units) of the largest rectangle

Question:

The area (in sq. units) of the largest rectangle $\mathrm{ABCD}$ whose vertices $\mathrm{A}$ and $\mathrm{B}$ lie on the $\mathrm{x}$-axis and vertices C and D lie on the parabola, $\mathrm{y}=\mathrm{x}^{2}-1$ below the $\mathrm{x}$-axis, is :

  1. $\frac{4}{3 \sqrt{3}}$

  2. $\frac{1}{3 \sqrt{3}}$

  3. $\frac{4}{3}$

  4. $\frac{2}{3 \sqrt{3}}$


Correct Option: 1

Solution:

Area $(\mathrm{A})=2 \mathrm{t} \cdot\left(1-\mathrm{t}^{2}\right)$

$(0<\mathrm{t}<1)$

$A=2 t-2 t^{3}$

$\frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}}=2-6 \mathrm{t}^{2}$

$t=\frac{1}{\sqrt{3}}$

$\Rightarrow A_{\max }=\frac{2}{\sqrt{3}}\left(1-\frac{1}{3}\right)=\frac{4}{3 \sqrt{3}}$

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