Question

The area of the rectangle which has the maximum perimeter
among all such rectangles, is?

Answer 1

Given : considering all rectangles lying in the region {(x, y) ∈ R × R : 0 ≤ x ≤ π/2 and 0 ≤ y ≤ 2sin2x} and having one side on the X - axis.

To find : The area of the rectangle which has the maximum parameter among all such rectangles, is ....

solution : let two points, (θ₁ , sin2θ₁) and (θ₂, sin2θ₂) lies on the curve y = 2sin2x such that these form a rectangle..

now from diagram it is clear that,

2sin2θ₁ = 2sin2θ₂

⇒2θ₁ = π - 2θ₂

⇒θ₁ = π/2 - θ₂......(1)

now parameter of rectangle, P(θ₁) = 2(length + width)

= 2[(θ₂ - θ₁) + 2sin2θ₁]

= 2[π/2 - 2θ₁ + 2sin2θ₁ ]

now differentiating P(θ₁) with respect to θ₁

so, P'(θ₁) = 2(-2 + 4cos2θ₁)

at P'(θ₁) = 0, θ₁ = π/6

again differentiating we get,

P"(θ₁) = (-8cos2θ₁) < 0

so, P(θ₁) will be maximum at θ₁ = π/6

now area of rectangle at θ₁ = π/6

area of rectangle = (θ₂ - θ₁) × 2sin2θ₁

= (π/2 - 2θ₁) × 2sin(π/3)

= (π/2 - π/3) × 2 × √3/2

= π/6 × √3

= π/2√3

Therefore the area of rectangle is π/2√3