A wire of length 36m 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?

Explanation

Let piece of wire used to make square be `l m `
then the length of the other piece will be `36−l`
Let from first piece we make the square, then

`x=ly`
`=>y=l/4`,where `y` is the side of the square .
`=>`Area of square
`=l^2/4^2`
From the second piece of length (36-l) we make a circle, then, `2pir=36-l`
`=>r=(36-l)/(2pi)`
`=>text(Area of circle )
=pi((36-l)/(2pi))^2`
Total area `= l^2/16+pi((36-l)/(2pi))^2`
`=l^2/16+(36-l)^2/(4pi)`
`=(dA)/(dl
)=l/8+(2(36-l)(-1))/(4pi)`
`=l/8-(36-l)/(2pi)`
differentiate again we get,
`(d^2A)/(dl^2)=1/8+1/(2pi)>0`
For` (dA)/(dl)=0`
`=>l/8-(36-l)/(2pi)=0`
`=>(pil-4(36-l))/(8pi)=0`
`=>(pi+4)l-144=0`
`=>l=144/(pi+4)`
Hence, the combined area is minimum when length of square is `l=144/(pi+4) m`
and that of circle is `(36-l)=(36pi)/(pi+4)m`