Use integration to find maximum area of rectangle inscribed in the ellipse
Answers
Answer:
Step-by-step explanation:
=> Equation for the ellipse:
x^2 / a^2 + y^2 / b^2 = 1
=> Suppose PQRS is a rectangle inscribed in the ellipse.
Coordinate for P = (acosፀ, bsinፀ)
Coordinate for S = (acosፀ, -bsinፀ)
Coordinate for Q = (-acosፀ, bsinፀ)
Coordinate for R = (-acosፀ, -bsinፀ)
=> Rectangle's Length:
RS = 2acosፀ
Rectangle's Breadth:
PS = 2bsinፀ
=> Let the area of rectangle = A
∴ A = RS * PS
A = 2acosፀ * 2bsinፀ
A = 4absinፀcosፀ
A = 2absin2ፀ
=> Differentiating both sides with respect to ፀ, we get
dA / dፀ = 2abcos 2ፀ * (2) = 4abcos2ፀ
dA/dፀ = 0 [For maximum and minimum]
4abcos2ፀ = 0
cos2ፀ = 0
2ፀ = π/2
ፀ = π/4
d^2A/dፀ^2 = -4absinፀ * (2) = -8absin2ፀ
When ፀ = π/4
d^2A/dፀ^2 = -8absin (2 * π/4) = -8absin π/4 = -8ab < 0
∴ point of max = π/4
If ፀ = π/4, then A is maximum
∴ A_max = 2absin(2*π/4) = 2absin π/2 = 2ab
Therefore, maximum area of rectangle inscribed in the ellipse is 2ab sq. units