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Worked Examples · Example 38

Q.A wire 40 m length 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 lengths of the two pieces so that the combined area of the square and the circle is minimum?

Puducherry CbseNCERTSubjective· 5mImportance★★★★★
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Let xx m form the square and (40−x)(40-x) m the circle. Total area A=x216+(40−x)24πA=\dfrac{x^2}{16}+\dfrac{(40-x)^2}{4\pi} is minimised at x=160π+4≈22.4x=\dfrac{160}{\pi+4}\approx22.4 m.

Square of perimeter xx: side =x4,=\dfrac{x}{4}, area =x216.=\dfrac{x^2}{16}. Circle of circumference c=2πrc=2\pi r: r=c2π,r=\dfrac{c}{2\pi}, area =πr2=c24π.=\pi r^2=\dfrac{c^2}{4\pi}. Minimise total area with A′(x)=0.A'(x)=0.

  1. Split the wire: let xx m be bent into a square and (40−x)(40-x) m into a circle, 0<x<40.0<x<40.
  2. Square area: side =x4⇒A1=(x4)2=x216.=\dfrac{x}{4}\Rightarrow A_1=\left(\dfrac{x}{4}\right)^2=\dfrac{x^2}{16}.
  3. Circle area: circumference =40−x=2πr⇒r=40−x2π;=40-x=2\pi r\Rightarrow r=\dfrac{40-x}{2\pi}; A2=πr2=(40−x)24π.A_2=\pi r^2=\dfrac{(40-x)^2}{4\pi}.
  4. Total area: A=x216+(40−x)24π.A=\dfrac{x^2}{16}+\dfrac{(40-x)^2}{4\pi}.
  5. Differentiate: A′(x)=2x16−2(40−x)4π=x8−40−x2π.A'(x)=\dfrac{2x}{16}-\dfrac{2(40-x)}{4\pi}=\dfrac{x}{8}-\dfrac{40-x}{2\pi}. …

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