Question

Use integration to find maximum area of rectangle inscribed in the ellipse

Answer 1

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