Question:
The maximum area (in sq. units) of a rectangle having its base on the x-axis and its other two vertices on the parabola, \(y = 12 - x^2\) such that the rectangle lies inside the parabola, is :
The maximum area (in sq. units) of a rectangle having its base on the x-axis and its other two vertices on the parabola, \(y = 12 - x^2\) such that the rectangle lies inside the parabola, is :
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Take the vertices as \((\pm x, 0)\) and \((\pm x, 12-x^2)\), write the area \(2x(12-x^2)\) and maximise it.
Updated On: Nov 5, 2026
- \(36\)
- \(20\sqrt{2}\)
- \(18\sqrt{3}\)
- \(32\)
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The Correct Option is D
Solution and Explanation
Let the rectangle have its base vertices at \((-x,0)\) and \((x,0)\); the upper vertices lie on the parabola at \((\pm x, 12-x^2)\).
Width = \(2x\), height = \(12-x^2\), so area \(A(x) = 2x(12-x^2) = 24x - 2x^3\).
\(A'(x) = 24 - 6x^2 = 0 \Rightarrow x^2 = 4 \Rightarrow x = 2\) (taking \(x \gt 0\)).
\(A''(x) = -12x \lt 0\) at \(x=2\), so this is a maximum.
\(A_{max} = 2(2)(12-4) = 4 \times 8 = 32\).
Answer: (4) 32
Width = \(2x\), height = \(12-x^2\), so area \(A(x) = 2x(12-x^2) = 24x - 2x^3\).
\(A'(x) = 24 - 6x^2 = 0 \Rightarrow x^2 = 4 \Rightarrow x = 2\) (taking \(x \gt 0\)).
\(A''(x) = -12x \lt 0\) at \(x=2\), so this is a maximum.
\(A_{max} = 2(2)(12-4) = 4 \times 8 = 32\).
Answer: (4) 32
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