# Ex 6.5,22 - Chapter 6 Class 12 Application of Derivatives (Term 1)

Last updated at April 15, 2021 by Teachoo

Ex 6.5

Ex 6.5, 1 (i)
Important

Ex 6.5, 1 (ii)

Ex 6.5, 1 (iii) Important

Ex 6.5, 1 (iv)

Ex 6.5, 2 (i)

Ex 6.5, 2 (ii) Important

Ex 6.5, 2 (iii)

Ex 6.5, 2 (iv) Important

Ex 6.5, 2 (v) Important

Ex 6.5, 3 (i)

Ex 6.5, 3 (ii)

Ex 6.5, 3 (iii)

Ex 6.5, 3 (iv) Important

Ex 6.5, 3 (v)

Ex 6.5, 3 (vi)

Ex 6.5, 3 (vii) Important

Ex 6.5, 3 (viii)

Ex 6.5, 4 (i)

Ex 6.5, 4 (ii) Important

Ex 6.5, 4 (iii)

Ex 6.5, 5 (i)

Ex 6.5, 5 (ii)

Ex 6.5, 5 (iii) Important

Ex 6.5, 5 (iv)

Ex 6.5,6

Ex 6.5,7 Important

Ex 6.5,8

Ex 6.5,9 Important

Ex 6.5,10

Ex 6.5,11 Important

Ex 6.5,12 Important

Ex 6.5,13

Ex 6.5,14 Important

Ex 6.5,15 Important

Ex 6.5,16

Ex 6.5,17

Ex 6.5,18 Important

Ex 6.5,19 Important

Ex 6.5, 20 Important

Ex 6.5,21

Ex 6.5,22 Important You are here

Ex 6.5,23 Important

Ex 6.5,24 Important

Ex 6.5,25 Important

Ex 6.5, 26 Important

Ex 6.5, 27 (MCQ)

Ex 6.5,28 (MCQ) Important

Ex 6.5,29 (MCQ)

Last updated at April 15, 2021 by Teachoo

Ex 6.5, 22 A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum? A wire of length 28m cut into two parts Let 𝑥 be the length of first part Length of 2nd part = (28−𝑥)𝑚 Given that one part is converted into a square Let length of 𝑥 m be converted into a square Thus, Perimeter of square =𝑥 4(𝑠𝑖𝑑𝑒 𝑜𝑓 𝑠𝑞𝑢𝑎𝑟𝑒 )=𝑥 side of square of =𝑥/4 Also, other part is converted into Circle ⇒ (28−𝑥) m length converted into a Circle of radius 𝑟 Circumference of circle = (28−𝑥) m 2𝜋𝑟=(28−𝑥) 𝑟=((28 − 𝑥)/2𝜋) m We need to find length of pieces so that the combined Area of Circle & Square is minimum Let T be the total area of circle & square T = Area of Circle + Area of Square T = 𝜋𝑟^2+(𝑠𝑖𝑑𝑒)^2 T = 𝜋((28 − 𝑥)/2𝜋)^2+𝑥^2/16 We have to minimize T Diff T w.r.t 𝑥 𝑑𝑇/𝑑𝑥=𝑑/𝑑𝑥 ((28 − 𝑥)^2/4𝜋+𝑥^2/16 " " ) = 2(28 − 𝑥)/4𝜋 . 𝑑(28 − 𝑥)/𝑑𝑥+2𝑥/16 = (28 − 𝑥)/2𝜋 (−1)+𝑥/8 Putting 𝒅𝑻/𝒅𝒙=𝟎 (𝑥 − 28)/2𝜋+𝑥/8=0 (4(𝑥 − 28) + 𝜋𝑥)/8𝜋 =0 4x – 112 +𝜋 𝑥=0×8 𝜋 𝑥(4+𝜋)=112 𝑥=112/(4 + 𝜋) Now finding (𝒅^𝟐 𝑻)/(𝒅𝒙^𝟐 ) 𝑑𝑇/𝑑𝑥=(−28 + 𝑥)/2𝜋+𝑥/8 (𝑑^2 𝑇)/(𝑑𝑥^2 )=1/2𝜋 (0+1)+1/8 (𝑑^2 𝑇)/(𝑑𝑥^2 )=1/2𝜋+1/8 > 0 Hence (𝑑^2 𝑇)/(𝑑𝑥^2 )>0 at 𝑥=112/(4 + 𝜋) ∴ Total area is Minimum at 𝑥 = 112/(4 + 𝜋) Finding length of other part Length of other part = 28 – x = 28 – 112/(4 + 𝜋) = (28(4 + 𝜋)−112)/(4 + 𝜋)= (28 𝜋)/(4 + 𝜋) ∴ Length of first part is 𝟏𝟏𝟐/(𝟒 + 𝝅) & another part is 𝟐𝟖𝝅/(𝟒 + 𝝅)