





Now learn Economics at Teachoo for Class 12
Minima/ maxima (statement questions) - Geometry questions
Ex 6.5, 27 (MCQ)
Example 35 Important
Example 41 Important
Example 36
Misc 12 Important
Example 37 Important
Misc 9 Important
Ex 6.5,21
Ex 6.5, 20 Important
Ex 6.5,24 Important
Ex 6.5,25 Important
Ex 6.5,26 Important
Ex 6.5,22 Important
Misc 10
Misc 11 Important
Ex 6.5,17
Ex 6.5,18 Important
Example 50 Important
Ex 6.5,19 Important You are here
Misc 8 Important
Ex 6.5,23 Important
Misc 15 Important
Misc 17 Important
Example 38 Important
Misc 18 Important
Minima/ maxima (statement questions) - Geometry questions
Last updated at April 15, 2021 by Teachoo
Ex 6.5, 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 (As AC is diameter of circle) 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, 𝑥 = 0 Which is not possible 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.