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Ex 6.5, 22 - A wire of length 28 m is to be cut into two pieces

Ex 6.5,22 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.5,22 - Chapter 6 Class 12 Application of Derivatives - Part 3
Ex 6.5,22 - Chapter 6 Class 12 Application of Derivatives - Part 4
Ex 6.5,22 - Chapter 6 Class 12 Application of Derivatives - Part 5
Ex 6.5,22 - Chapter 6 Class 12 Application of Derivatives - Part 6


Transcript

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 π‘₯ be the length of first part Length of 2nd part = (28βˆ’π‘₯)π‘š Given that one part is converted into a square Let length of π‘₯ m be converted into a square Thus, Perimeter of square =π‘₯ 4(𝑠𝑖𝑑𝑒 π‘œπ‘“ π‘ π‘žπ‘’π‘Žπ‘Ÿπ‘’ )=π‘₯ side of square of =π‘₯/4 Also, other part is converted into Circle β‡’ (28βˆ’π‘₯) m length converted into a Circle of radius π‘Ÿ Circumference of circle = (28βˆ’π‘₯) m 2πœ‹π‘Ÿ=(28βˆ’π‘₯) π‘Ÿ=((28 βˆ’ π‘₯)/2πœ‹) 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 = πœ‹π‘Ÿ^2+(𝑠𝑖𝑑𝑒)^2 T = πœ‹((28 βˆ’ π‘₯)/2πœ‹)^2+π‘₯^2/16 We have to minimize T Diff T w.r.t π‘₯ 𝑑𝑇/𝑑π‘₯=𝑑/𝑑π‘₯ ((28 βˆ’ π‘₯)^2/4πœ‹+π‘₯^2/16 " " ) = 2(28 βˆ’ π‘₯)/4πœ‹ . 𝑑(28 βˆ’ π‘₯)/𝑑π‘₯+2π‘₯/16 = (28 βˆ’ π‘₯)/2πœ‹ (βˆ’1)+π‘₯/8 Putting 𝒅𝑻/𝒅𝒙=𝟎 (π‘₯ βˆ’ 28)/2πœ‹+π‘₯/8=0 (4(π‘₯ βˆ’ 28) + πœ‹π‘₯)/8πœ‹ =0 4x – 112 +πœ‹ π‘₯=0Γ—8 πœ‹ π‘₯(4+πœ‹)=112 π‘₯=112/(4 + πœ‹) Now finding (𝒅^𝟐 𝑻)/(𝒅𝒙^𝟐 ) 𝑑𝑇/𝑑π‘₯=(βˆ’28 + π‘₯)/2πœ‹+π‘₯/8 (𝑑^2 𝑇)/(𝑑π‘₯^2 )=1/2πœ‹ (0+1)+1/8 (𝑑^2 𝑇)/(𝑑π‘₯^2 )=1/2πœ‹+1/8 > 0 Hence (𝑑^2 𝑇)/(𝑑π‘₯^2 )>0 at π‘₯=112/(4 + πœ‹) ∴ Total area is Minimum at π‘₯ = 112/(4 + πœ‹) Finding length of other part Length of other part = 28 – x = 28 – 112/(4 + πœ‹) = (28(4 + πœ‹)βˆ’112)/(4 + πœ‹)= (28 πœ‹)/(4 + πœ‹) ∴ Length of first part is 𝟏𝟏𝟐/(πŸ’ + 𝝅) & another part is πŸπŸ–π…/(πŸ’ + 𝝅)

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.