MHT CET · Maths · Application of Derivatives
A rectangle of maximum area is inscribed in an ellipse \(\frac{x^2}{25}+\frac{y^2}{16}=1\), then its dimensions are
- A \(4 \sqrt{2}, 6 \sqrt{2}\)
- B \(\sqrt{2}, 5 \sqrt{2}\)
- C \(4 \sqrt{2}, 5 \sqrt{2}\)
- D \(4 \sqrt{2}, \sqrt{2}\)
Answer & Solution
Correct Answer
(C) \(4 \sqrt{2}, 5 \sqrt{2}\)
Step-by-step Solution
Detailed explanation

Length of rectangle \(=10 \cos \theta\) and
breadth of rectangle \(=8 \sin \theta\)
\(\therefore\) Area of rectangle \(=(10 \cos \theta)(8 \sin \theta)=40(\sin \theta)\)
Maximum area will occur when \(\sin 2 \theta=1\)
\(\therefore \sin 2 \theta=\sin \frac{\pi}{2} \quad \Rightarrow \theta=\frac{\pi}{4} \)
\( \therefore P=\left(\frac{5}{\sqrt{2}}, \frac{4}{\sqrt{2}}\right) \Rightarrow \text { Dimensions of rectangle are }\) \(5 \sqrt{2}, 4 \sqrt{2}\)
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