Question

A wire of length 20 m is to the cut into two pieces. A piece of length $l_1$ is bent to make a square of area $A_1$ and the other piece of length $l_2$ is made into a circle of area $A_2$. If $2A_1 +3A_2$ is minimum then $(\pi l_1): l_2$ is equal to:

• A. 6:1
• B. 3:1
• C. 4:1
• D. 1:6

Answer 1

Answer: (A)

6:1

Answer 2

Let $l_1$ and $l_2$ be the lengths of the two pieces, so $l_1 + l_2 = 20$. The area of the square is $A_1 = (\frac{l_1}{4})^2 = \frac{l_1^2}{16}$. The area of the circle is $A_2 = \pi (\frac{l_2}{2\pi})^2 = \frac{l_2^2}{4\pi}$. We want to minimize $E = 2A_1 + 3A_2 = \frac{l_1^2}{8} + \frac{3l_2^2}{4\pi}$. Substitute $l_2 = 20 - l_1$: $E(l_1) = \frac{l_1^2}{8} + \frac{3(20 - l_1)^2}{4\pi}$. To find the minimum, set $\frac{dE}{dl_1} = \frac{l_1}{4} - \frac{3(20 - l_1)}{2\pi} = 0$. This gives $\frac{l_1}{4} = \frac{3(20 - l_1)}{2\pi}$, which simplifies to $2\pi l_1 = 12(20 - l_1)$. Solving for $l_1$: $2\pi l_1 = 240 - 12l_1 \implies (2\pi + 12)l_1 = 240 \implies l_1 = \frac{120}{\pi + 6}$. Then $l_2 = 20 - l_1 = 20 - \frac{120}{\pi + 6} = \frac{20\pi}{\pi + 6}$. The ratio $(\pi l_1) : l_2 = (\pi \frac{120}{\pi + 6}) : (\frac{20\pi}{\pi + 6}) = 120\pi : 20\pi = 6:1$.