Consider all rectangles lying in the region x ,   y ∈ R × R : 0 ≤ x ≤ π 2   a n d   0 ≤ y ≤ 2 sin 2 x and having one side on the x -axis. The area of the rectangle which has the maximum perimeter among all such rectangles, is
C. π 2 3
2 sin 2 θ 1 = 2 sin 2 θ 2 ⇒ 2 θ 1 = π − 2 θ 2 ⇒ θ 2 = π 2 − θ 1 … . ( 1 ) Now perimeter p θ 1 , θ 2 = 2 θ 2 − θ 1 + 2 sin 2 θ 1 ⇒ p θ 1 = 2 π 2 − 2 θ 1 + 2 sin 2 θ 1 ⇒ p ' θ 1 = 2 − 2 + 4 cos 2 θ 1 Also, p " θ 1 = 2 − 8 sin 2 θ 1 0 For maximum perimeter, p ' θ 1 = 0 & p " θ 1 0 ⇒ cos 2 θ 1 = 1 2 ⇒ 2 θ 1 = π 3 ⇒ θ 1 = π 6 Now area at θ 1 = π 6 will be θ 2 − θ 1 × 2 sin 2 θ 1 = π 2 − 2 θ 1 ⋅ 2 sin 2 θ 1 = π 2 − π 3 × 2 sin π 3 = π 6 ⋅ 3 = π 2 3
Related: Mathematics — Application of Derivatives · All PYQ Banks