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 $28-\frac{112}{\pi+4}=\frac{28\pi}{\pi+4} cm.$