A wire of length 2 units is cut into two parts which are bent respectively to form a square of side = x units and a circle of radius = r units. If the sum of the areas of the square and the circle so formed is minimum, then
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Answer:2r
Explanation:Solution :
Let length of two parts be ‘a’ and ‘2 - a’
As per condition given, we write
a=4xand2−a=2πr
x=a4 and r=2−a2π
∴A(square)=(a4)2=a216 and
A(circle)=π[(2−a)2π]2=π(4+a2−4a)4π2
=(a2−4a+4)4π
f(a)=a216+a2−4a+44π
∴f(a)=a2π+4a2−16a+1616π
f′(a)=116π[2aπ+8a−16]
f′(a)=0=>2aπ+8a−16=0
=> 2aπ+8a=16
x=a4=2π+4
and r=2−a2π
=2−8π+42π
=2π+8−82π(π+4)
=1π+4
x=2π+4 and r=1π+4
x=2r
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