A rectangle has two of its vertices on the x-axis and the other two vertices on the parabola \(y = 12 - x^{2}\), above the x-axis. The maximum possible area of the rectangle is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 32 square units
The top two vertices are at the same height, and the parabola is symmetric about the y-axis, so they are \((\pm x, 12 - x^{2})\) with \(0 \lt x \lt 2\sqrt{3}\). The area is
\( A = 2x(12 - x^{2}) = 24x - 2x^{3} \)
Setting the derivative equal to zero:
\( A' = 24 - 6x^{2} = 0,\ x = 2 \)
Since \(A'' = -12x = -24 \lt 0\) at x = 2, this is the maximum:
\( A = 2 \times 2 \times (12 - 4) = 32 \) square units