Using integration, find the area of greatest rectangle that can be inscribed in a ellipse
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Step-by-step explanation:
=> Area, A = ()(2acosФ)(2bsinФ) = 2absin^2Ф
dA/dФ = 4abcos2Ф
Area is maximum or minimum
when, dA/dФ = 0
=> cos2Ф = 0
=> 2Ф = π/2
=> Ф = π/4
d^2A/dФ^2 = -8absin^2Ф
d^2A/dФ^2 )_Ф=π/4 = -8absinπ/2 = -8ab < 0
=> The area of greatest rectangle that can be inscribed in a ellipse
∴ Area is maximum when Ф = π/4
=> A_max = 2absin2Ф)_Ф = π/4
2absinπ/2 = 2ab
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