Question

Find the area of the greatest rectangle that can be inscribed in an ellipse \[\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\]

Answer 1

Hint: We can assume a rectangle inside an ellipse now we know that general coordinates of any point is \[(acos\theta ,bsin\theta )\] , so we consider 4 vertices of rectangles as \[(acos\theta ,bsin\theta ), (acos\theta ,-bsin\theta ), (-acos\theta ,bsin\theta ) and (-acos\theta ,-bsin\theta )\]
So side length of rectangle will be \[2acos\theta \] and \[2bsin\theta \] so its area will be \[ab4sin\theta cos\theta \]
Now we have to maximize area by differentiating and equating 0, and double differentiating and equating < 0.

Complete step-by-step solution:
We are given an ellipse \[\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\] and a rectangle is inscribed in it so as we know general coordinates of any point in ellipse of equation \[\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\] is \[(acos\theta ,bsin\theta )\]

So, we take general vertices of rectangle as \[(acos\theta, bsin\theta ), (acos\theta, -bsin\theta ), (-acos\theta, bsin\theta ) and (-acos\theta, -bsin\theta )\]
So, we can see that length and breadth of rectangle is \[2acos\theta \] and \[2bsin\theta \]
So, its area will be \[ab4sin\theta cos\theta \] (length x Breadth)
Applying trigonometric formula \[2sin\theta cos\theta =\sin 2\theta \], let area equals to A
Now \[A=2ab\sin 2\theta \]
Now as its the area is maximum given in question, we have to differentiate it with respect to \[\theta \]
\[\dfrac{dA}{d\theta }=2ab\cos 2\theta \times 2\] (using \[\dfrac{d\sin 2\theta }{d\theta }=2\cos 2\theta \])
Equating \[\dfrac{dA}{d\theta }=2ab\cos 2\theta \times 2=0\], it means \[\cos 2\theta =0\] , which means \[2\theta =\dfrac{\pi }{2}\] and \[\theta =\dfrac{\pi }{4}\]
But we have to check that at \[\theta =\dfrac{\pi }{4}\] area is maximum and not minimum
For that we have to calculate \[\dfrac{{{d}^{2}}A}{d{{\theta }^{2}}}\]
Which will be equals to \[\dfrac{{{d}^{2}}A}{d{{\theta }^{2}}}=4ab(-\sin 2\theta )\times 2\]
On solving further equals to \[\dfrac{{{d}^{2}}A}{d{{\theta }^{2}}}=-8ab(\sin 2\theta )\] now putting \[\theta =\dfrac{\pi }{4}\] we get
\[\dfrac{{{d}^{2}}A}{d{{\theta }^{2}}}=-8ab\], which is a negative quantity hence we can say that area is maximum at \[\theta =\dfrac{\pi }{4}\]
So now we to put \[\theta =\dfrac{\pi }{4}\] in equation of area which is \[A=2ab\sin 2\theta \]
We get as \[A=2ab\sin 2\dfrac{\pi }{4}=2ab\sin \dfrac{\pi }{2}=2ab\]
Hence max area of rectangle inside ellipse is \[2ab\].

Note: If instead of ellipse we are given a circle \[{{x}^{2}}+{{y}^{2}}={{r}^{2}}\] and to find area of the greatest rectangle that can be inscribed in it , again we can apply same procedure but this time the coordinates are \[(rcos\theta ,rsin\theta ),(rcos\theta ,-rsin\theta ),(-rcos\theta ,rsin\theta )and(-rcos\theta ,-rsin\theta )\]
So, we can see that length and breadth of the rectangle is \[2rcos\theta \] and \[2rsin\theta \]. So, its area will be \[{{r}^{2}}4sin\theta cos\theta \] (length x Breadth)
And after differentiating it results into \[2{{r}^{2}}\]