A) \[\sqrt{2}\]
B) \[2\]
C) \[\sqrt{3}\]
D) \[\frac{\sqrt{2}}{3}\]
E) \[\frac{-2}{\sqrt{3}}\]
Correct Answer: A
Solution :
Let the length and breadth of the rectangle be \[l\]and b. In\[\Delta OAB,\] \[b=\sin \theta \] and \[l=2\cos \theta \] \[\therefore \] A = Area of rectangle \[=lb=2\text{ }sin\theta \text{ }cos\theta \] \[=sin2\theta \] On differentiating w.r.t.\[\theta \], we get \[\frac{dA}{d\theta }=2\cos 2\theta \] Put\[\frac{dA}{d\theta }=0\]for maxima or minima \[\Rightarrow \] \[\cos 2\theta =0\] \[\Rightarrow \] \[\theta =\frac{\pi }{4}\] \[\therefore \] \[\frac{{{d}^{2}}A}{d{{\theta }^{2}}}=-4\sin 2\theta \] \[{{\left( \frac{{{d}^{2}}A}{d{{\theta }^{2}}} \right)}_{\theta =\frac{\pi }{4}}}<0\] \[\therefore \]Function is maximum at\[\theta =\frac{\pi }{4}\]. \[\therefore \]Length of rectangle\[=2\cos \theta =2\cos \frac{\pi }{4}\] \[=2\frac{1}{\sqrt{2}}=\sqrt{2}\]You need to login to perform this action.
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