Ex 6.3

Chapter 6 Class 12 Application of Derivatives
Serial order wise

### Transcript

Ex 6.3, 19 Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area.Let radius be r of the circle & let 𝑥 be the length & 𝑦 be the breadth of the rectangle Now, Δ ABC is right angle triangle (AB)2 + (BC)2 = (AC)2 𝑥^2+𝑦^2 = (2𝑟)^2 𝑥^2+𝑦^2= 4𝑟2 𝑦2 = 4𝑟2 – 𝑥2 𝑦= √(4𝑟"2 – " 𝑥"2" ) We need to maximize Area of rectangle Let A be the area rectangle Area of rectangle = Length × Breadth A = xy A = 𝑥 √(4𝑟^2−𝑥^2 ) Since A has square root It will be difficult to differentiate So, we take Z = A2 Let Z = A2 Z = 𝑥^2× (√(4𝑟^2−𝑥^2 ))^2 Z = 𝑥^2×(4𝑟^2−𝑥^2 ) Z = 4𝑟^2 𝑥^2−𝑥^4 Since A is positive, A is maximum if A2 is maximum So, we maximize Z = A2 Diff. Z w.r.t 𝑥 𝑑Z/𝑑𝑥=𝑑(4𝑟^2 𝑥^2−𝑥^4 )/𝑑𝑥 𝑑Z/𝑑𝑥=4𝑟^2×2𝑥−4𝑥^3 𝑑Z/𝑑𝑥=8𝑟^2 𝑥−4𝑥^3 Putting 𝑑Z/𝑑𝑥 = 0 8𝑟^2 𝑥−4𝑥^3 = 0 4𝑥^3−8𝑟^2 𝑥 = 0 𝑥^3−2𝑟^2 𝑥 = 0 𝑥 (𝑥^2−2𝑟^2 ) = 0 Therefore, Finding (𝒅^𝟐 𝐙)/(𝐝𝒙^𝟐 ) 𝑑Z/𝑑𝑥=8𝑟^2 𝑥−4𝑥^3 Diff w.r.t 𝑥 (𝑑^2 Z)/(𝑑𝑥^2 ) = 𝑑/𝑑𝑥 [8𝑟^2 𝑥−4𝑥^3 ] (𝑑^2 Z)/(𝑑𝑥^2 ) = 8𝑟^2−4×3𝑥^2 (𝑑^2 Z)/(𝑑𝑥^2 ) = 8𝑟^2−12𝑥^2 Putting 𝒙^𝟐=𝟐𝒓^𝟐 (𝑑^2 Z)/(𝑑𝑥^2 ) = 8𝑟^2−12×2𝑟^2 (𝑑^2 Z)/(𝑑𝑥^2 ) = 8𝑟^2−24𝑟^2 (𝑑^2 Z)/(𝑑𝑥^2 ) = −16𝑟^2 < 0 Hence, (𝑑^2 Z)/(𝑑𝑥^2 ) < 0 at 𝑥^2=2𝑟^2 Thus area is maximum when 𝑥^2=2𝑟^2 Now, finding y 𝑦 = √(4𝑟^2 −𝑥^2 ) Putting 𝑥^2=2𝑟^2 𝑦 = √(4𝑟^2 −〖2𝑟〗^2 ) 𝑦 = √(〖2𝑟〗^2 ) 𝑦 = √2 𝑟 Therefore 𝑥 = 𝑦 = √2 𝑟 Hence area is maximum when 𝒙 = 𝒚 ∴ The rectangle is a square.

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.