Question

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

• A. \(\frac{1}{3\sqrt{3}}\)
• B. \(\frac{2}{3\sqrt{3}}\)
• C. \(\frac{4}{3\sqrt{3}}\)
• D. \(\frac{4}{3}\)

Hint:

Take C = \((a, a^2-1)\) and D = \((-a, a^2-1)\); the rectangle has width 2a and height \(1-a^2\). Maximise the area.

Answer 1

The Correct Option is C

Let C = \((a, a^2-1)\) and D = \((-a, a^2-1)\) with \(0 \lt a \lt 1\) (below the x-axis). Then A = \((-a,0)\), B = \((a,0)\).
Width = \(2a\), height = \(1-a^2\).
Area \(S(a) = 2a(1-a^2) = 2a - 2a^3\).
\(S'(a) = 2 - 6a^2 = 0 \Rightarrow a = \frac{1}{\sqrt{3}}\); \(S''(a) = -12a \lt 0\), so this is a maximum.
\(S_{max} = 2\cdot\frac{1}{\sqrt3}\left(1-\frac13\right) = \frac{2}{\sqrt3}\cdot\frac23 = \frac{4}{3\sqrt3}\).
Answer: (3) \(\frac{4}{3\sqrt{3}}\)