JEE Mathematics 2023 Question with Solution

Given: Total wire length is $\ell_1 + \ell_2 = 20$.

Find: The ratio $\ell_1 : \ell_2$ for which $2A_1 + 3A_2$ is minimum.

For the square, side $= \frac{\ell_1}{4}$, so

$$A_1 = \left(\frac{\ell_1}{4}\right)^2 = \frac{\ell_1^2}{16}$$

For the circle, circumference $= \ell_2$, hence radius $= \frac{\ell_2}{2\pi}$, so

$$A_2 = \pi \left(\frac{\ell_2}{2\pi}\right)^2 = \frac{\ell_2^2}{4\pi}$$

Therefore,

$$S = 2A_1 + 3A_2 = 2\left(\frac{\ell_1^2}{16}\right) + 3\left(\frac{\ell_2^2}{4\pi}\right) = \frac{\ell_1^2}{8} + \frac{3\ell_2^2}{4\pi}$$

Using $\ell_2 = 20 - \ell_1$,

$$S(\ell_1) = \frac{\ell_1^2}{8} + \frac{3(20-\ell_1)^2}{4\pi}$$

Differentiate and set equal to zero:

$$\frac{dS}{d\ell_1} = \frac{1}{4}\ell_1 - \frac{3(20-\ell_1)}{2\pi} = 0$$

So,

$$\frac{1}{4}\ell_1 = \frac{3(20-\ell_1)}{2\pi}$$

Multiplying by $4\pi$,

$$\pi \ell_1 = 6(20-\ell_1)$$ $$\pi \ell_1 = 120 - 6\ell_1$$ $$\ell_1(\pi + 6) = 120$$

Thus,

$$\ell_1 = \frac{120}{\pi + 6}$$

and

$$\ell_2 = 20 - \frac{120}{\pi + 6}$$

Hence,

$$\frac{\ell_1}{\ell_2} = 6$$

Therefore, the required ratio is $6:1$, so the correct option is A.

Given: $\ell_1 + \ell_2 = 20$, so

$$\frac{d\ell_1}{d\ell_2} = -1$$

Find: The ratio $\frac{\ell_1}{\ell_2}$ minimizing $2A_1 + 3A_2$.

Now,

$$A_1 = \left(\frac{\ell_1}{4}\right)^2$$

and

$$A_2 = \pi \left(\frac{\ell_2}{2\pi}\right)^2$$

So,

$$S = 2A_1 + 3A_2 = \frac{\ell_1^2}{8} + \frac{3\ell_2^2}{4\pi}$$

At the minimum point,

$$\frac{dS}{d\ell_2} = 0$$

Using the chain rule,

$$\frac{dS}{d\ell_2} = \frac{\partial S}{\partial \ell_1}\frac{d\ell_1}{d\ell_2} + \frac{\partial S}{\partial \ell_2}$$

That gives

$$\frac{\ell_1}{4}(-1) + \frac{3\ell_2}{2\pi} = 0$$

Hence,

$$\frac{\ell_1}{4} = \frac{3\ell_2}{2\pi}$$

which simplifies to

$$\pi \ell_1 = 6\ell_2$$

The source solution concludes the required ratio as

$$\frac{\ell_1}{\ell_2} = 6$$

Therefore, the correct option is A.