A wire of length 20 m is to be cut into two pieces. A piece of length ℓ₁ is bent to make a square of

Q: Wait, why do we reject the negative root here? Or do we even have a quadratic?

Great question! We don't actually have a quadratic equation to solve for $\ell_1$ and $\ell_2$ individually. Because the problem asks for a ratio, we are performing constrained optimization. The derivative $\frac{dS}{d\ell_1} = 0$ gives us a linear relationship between $\ell_1$ and $\ell_2$, which is exactly what we need for the ratio.

Q: Why is the derivative of $\ell_2$ with respect to $\ell_1$ equal to $-1$?

Remember our constraint: $\ell_1 + \ell_2 = 20$. If you differentiate both sides with respect to $\ell_1$, you get $\frac{d}{d\ell_1}(\ell_1) + \frac{d}{d\ell_1}(\ell_2) = \frac{d}{d\ell_1}(20)$. This simplifies to $1 + \frac{d\ell_2}{d\ell_1} = 0$, which means $\frac{d\ell_2}{d\ell_1} = -1$. It's a fundamental step in handling linked variables!

Q: Why can't we just use energy conservation or some other physics principle?

This is a pure optimization problem, not a physical system in motion. In JEE Advanced, you'll often encounter 'optimization' problems disguised as geometry or physics. The 'energy' here is the cost function $S$, and we are finding the state of minimum 'potential' (or minimum area sum) by using calculus.

Question

A wire of length $20	ext{ m}$ is to be cut into two pieces. A piece of length $\ell_1$ is bent to make a square of area $A_1$ and the other piece of length $\ell_2$ is made into a circle of area $A_2$. If $2A_1 + 3A_2$ is minimum then $(\pi \ell_1) : \ell_2$ is equal to:

Options:

(A) $6 : 1$

(B) $3 : 1$

(C) $1 : 6$

(D) $4 : 1$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a twenty-meter wire, holding a pair of shears. You are tasked with a challenge: cut this wire into two pieces, $\ell_1$ and $\ell_2$, and transform them into a square and a circle, respectively.

Your goal is to minimize the sum of their areas, defined by the function $S = 2A_1 + 3A_2$. This is a classic optimization problem involving geometry and calculus.

### The Geometry of the Pieces

First, we translate the physical reality into algebraic language. Given a total length of $20	ext{ m}$, our constraint is:
$$\ell_1 + \ell_2 = 20$$

For the square, if we use length $\ell_1$ to form the perimeter, each side is $\frac{\ell_1}{4}$. The area $A_1$ is therefore:
$$A_1 = \left( \frac{\ell_1}{4} ight)^2 = \frac{\ell_1^2}{16}$$

For the circle, using length $\ell_2$ as the circumference, we have $2\pi r = \ell_2$, which gives a radius $r = \frac{\ell_2}{2\pi}$. The area $A_2$ is:
$$A_2 = \pi r^2 = \pi \left( \frac{\ell_2}{2\pi} ight)^2 = \frac{\ell_2^2}{4\pi}$$

### The Objective Function

We construct our objective function $S = 2A_1 + 3A_2$. Substituting our geometric expressions, we obtain:
$$S = 2\left( \frac{\ell_1^2}{16} ight) + 3\left( \frac{\ell_2^2}{4\pi} ight)$$

Simplifying this, we arrive at the function we must minimize:
$$S = \frac{\ell_1^2}{8} + \frac{3\ell_2^2}{4\pi}$$

### The Calculus of Change

To find the minimum, we differentiate $S$ with respect to $\ell_1$. Since $\ell_1 + \ell_2 = 20$, we know that $\frac{d\ell_2}{d\ell_1} = -1$.

Differentiating $S$ with respect to $\ell_1$ using the chain rule:
$$\frac{dS}{d\ell_1} = \frac{d}{d\ell_1} \left( \frac{\ell_1^2}{8} ight) + \frac{d}{d\ell_1} \left( \frac{3\ell_2^2}{4\pi} ight)$$

Applying the power rule and the chain rule, we get:
$$\frac{dS}{d\ell_1} = \frac{2\ell_1}{8} + \frac{6\ell_2}{4\pi} \cdot \left( \frac{d\ell_2}{d\ell_1} ight)$$

Substituting $\frac{d\ell_2}{d\ell_1} = -1$, the derivative becomes:
$$\frac{dS}{d\ell_1} = \frac{\ell_1}{4} - \frac{6\ell_2}{4\pi}$$

### The Elegant Conclusion

For the minimum, we set the derivative to zero:
$$\frac{\ell_1}{4} - \frac{6\ell_2}{4\pi} = 0$$

This simplifies to:
$$\frac{\ell_1}{4} = \frac{6\ell_2}{4\pi}$$

Canceling the $4$ from both denominators, we are left with $\ell_1 = \frac{6\ell_2}{\pi}$. Rearranging this, we find $\pi \ell_1 = 6\ell_2$.

The ratio $\pi \ell_1 : \ell_2$ is **$6 : 1$**. We have navigated the geometry, mastered the calculus, and arrived at the elegant solution.

Select Answer:

Visualized Solution

The Wire and the Cut

Geometry of the Square

Geometry of the Circle

The Objective Function

Handling the Variables

Differentiating the Objective Function

Applying the Chain Rule Result

Finding the Critical Point

Solving for the Ratio

Final Conclusion

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Geometry of the Pieces

The Objective Function

The Calculus of Change

The Elegant Conclusion

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Select Answer:

Visualized Solution

The Wire and the Cut

Geometry of the Square

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The Objective Function

Handling the Variables

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Applying the Chain Rule Result

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Final Conclusion

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Similar Questions

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If a rectangle is inscribed in an equilateral triangle of side length $22$ as shown in the figure, then the square of the largest area of such a rectangle is ___