Question

A wire of length 28m 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?

Answer 1

Let a piece of length l be cut from the given wire to make a square.

Then,the other piece of wire to be made into a circle is of length\((28−l) m.\)

Now,side of square =\(\frac{l}{4}\).

Let r be the radius of the circle.Then,\(2\pi r=28-l\)⇒\(r=\frac{1}{2\pi}(28-l).\)

The combined areas of the square and the circle\((A)\)is given by,

\(A=(side \space of\space  the \space square)\pi ^{{2}}+r^{2}\)

\(=\frac{l^{2}}{16}+\pi[\frac{1}{2\pi}(28-l)]^{2}\)

\(=\frac{l^{2}}{16}+\frac{1}{4\pi}(28-l)^{2}\)

\(∴\frac{dA}{dl}\)=\(\frac{2l}{16}+\frac{2}{4\pi }(28-l)(-1)\)\(=\frac{l}{8}-\frac{1}{2\pi}(28-l)\)

\(\frac{d^{2}A}{dl^{2}}=\frac{1}{8}+\frac{1}{2\pi }>0\)

Now,\(\frac{dA}{dl}=0⇒\frac{l}{8}-\frac{1}{2\pi}(28-l)=0\)

\(⇒\frac{\pi l-4(28-l)}{8\pi}=0\)

\(⇒(\pi +4)l-112=0\)

\(⇒l=\frac{112}{\pi+4}\)

Thus,when\(⇒l=\frac{112}{\pi+4}\),\(\frac{d^{2}A}{dl^{2}}>0\).

By second derivative test,the area(A)is the minimum when \(l=\frac{112}{\pi+4}\)

Hence,the combined area is the minimum when the length of the wire in making the

square is \(l=\frac{112}{\pi+4}\) cm while the length of the wire in making the circle

is \(28-\frac{112}{\pi+4}=\frac{28\pi}{\pi+4} cm.\)