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Question

A wire of length l is cut into two parts. One part is bent into a circle and other into a square. Show that the sum of areas of the circle and square is the least, if the radius of circle is half the side of the square

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Solution

Given, length of wire is l.
Let length of one part be x, then other part be lx.
Let x part be bent into a circle and lx part be in a square.
2πr=x and 4s=1xs=lx4(s= side )
According to question, we have
r=s2=lx8
Also, r=x2π
lx8=x2π
l=8x2π+x=4xπ+x
Sum of areas of circle and square,
f(x)=π(lx8)2+(lx4)2
f(x)=π64[(lx)2+4(lx)2]
f(x)=π645(lx)2
=5π64(4xπ+xx)2
f(x)=5π6416x2π2
f(x)=5x24π
f(x)=54π2x=5x2π
For minimum value, f(x)=0
52πx=0
x=0
If x=of"(0)=52π>0
f(x) attains minimum at x=0.

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