JEE Mains · Maths · STD 12 - 6. Application of derivatives
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
- A \(36\)
- B \(20\sqrt 2 \)
- C \(32\)
- D \(18\sqrt 3 \)
Answer & Solution
Correct Answer
(C) \(32\)
Step-by-step Solution
Detailed explanation
\(A=2 \alpha\left(12-\alpha^{2}\right)\) \(\frac{{dA}}{{d\alpha }} = 0\) \( \Rightarrow 2\left( {12 - 3{\alpha ^2}} \right) = 0\) \( \Rightarrow \alpha = \pm 2\) \(A=4(12-4)=32\)
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