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# A wire of length 28 m is to be cut into two pieces.

Question:

A wire of length 28 m 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?

Solution:

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, $2 \pi r=28-l \Rightarrow r=\frac{1}{2 \pi}(28-l)$.

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

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

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

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

$\therefore \frac{d A}{d l}=\frac{2 l}{16}+\frac{2}{4 \pi}(28-l)(-1)=\frac{l}{8}-\frac{1}{2 \pi}(28-l)$

$\frac{d^{2} A}{d l^{2}}=\frac{1}{8}+\frac{1}{2 \pi}>0$

Now, $\frac{d A}{d l}=0 \Rightarrow \frac{l}{8}-\frac{1}{2 \pi}(28-l)=0$

$\Rightarrow \frac{\pi l-4(28-l)}{8 \pi}=0$

$\Rightarrow(\pi+4 l-112=0$

$\Rightarrow l=\frac{112}{\pi+4}$

Thus, when $l=\frac{112}{\pi+4}, \frac{d^{2} \mathrm{~A}}{d l^{2}}>0 .$

$\therefore$ 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 $\frac{112}{\pi+4} \mathrm{~m}$ while the length of the wire in making the circle is

$28-\frac{112}{\pi+4}=\frac{28 \pi}{\pi+4} \mathrm{~m}$