Key Concepts and Formulas

• Optimization using Derivatives: Finding the minimum or maximum value of a function by finding critical points (where the derivative is zero) and using the second derivative test to determine the nature of the critical point.
• Area of a Square: If $s$ is the side length of a square, its area is $A = s^2$.
• Area and Circumference of a Circle: If $r$ is the radius of a circle, its area is $A = \pi r^2$ and its circumference is $C = 2\pi r$.

Step-by-Step Solution

1. Define Variables and Formulate the Constraint

Let $l_1$ be the length of the wire used to form the square, and $l_2$ be the length of the wire used to form the circle. The total length of the wire is $20$ m. Therefore, the constraint is:

$l_1 + l_2 = 20$

This constraint will be used to eliminate one variable and express the objective function in terms of a single variable.

2. Express Areas $A_1$ and $A_2$ in terms of $l_1$ and $l_2$

• Area of the Square ($A_1$): The perimeter of the square is $l_1$. If $s$ is the side length of the square, then $4s = l_1$, so $s = \frac{l_1}{4}$. The area of the square is $A_1 = s^2$. $A_1 = \left(\frac{l_1}{4}\right)^2 = \frac{l_1^2}{16}$

• Area of the Circle ($A_2$): The circumference of the circle is $l_2$. If $r$ is the radius of the circle, then $2\pi r = l_2$, so $r = \frac{l_2}{2\pi}$. The area of the circle is $A_2 = \pi r^2$. $A_2 = \pi \left(\frac{l_2}{2\pi}\right)^2 = \pi \left(\frac{l_2^2}{4\pi^2}\right) = \frac{l_2^2}{4\pi}$

3. Formulate the Objective Function

We want to minimize the quantity $S = 2A_1 + 3A_2$. Substituting the expressions for $A_1$ and $A_2$, we get:

$S = 2\left(\frac{l_1^2}{16}\right) + 3\left(\frac{l_2^2}{4\pi}\right) = \frac{l_1^2}{8} + \frac{3l_2^2}{4\pi}$

4. Express the Objective Function in Terms of a Single Variable

Using the constraint $l_1 + l_2 = 20$, we can express $l_2$ in terms of $l_1$ as $l_2 = 20 - l_1$. Substituting this into the expression for $S$, we have:

$S(l_1) = \frac{l_1^2}{8} + \frac{3(20 - l_1)^2}{4\pi}$

5. Find the Critical Points by Differentiation

To find the minimum value of $S(l_1)$, we differentiate $S(l_1)$ with respect to $l_1$ and set the derivative equal to zero:

$\frac{dS}{dl_1} = \frac{d}{dl_1}\left(\frac{l_1^2}{8} + \frac{3(20 - l_1)^2}{4\pi}\right) = \frac{2l_1}{8} + \frac{3(2)(20 - l_1)(-1)}{4\pi} = \frac{l_1}{4} - \frac{3(20 - l_1)}{2\pi}$

Setting $\frac{dS}{dl_1} = 0$:

$\frac{l_1}{4} - \frac{3(20 - l_1)}{2\pi} = 0$

6. Solve for the Relationship between $l_1$ and $l_2$

From the previous step:

$\frac{l_1}{4} = \frac{3(20 - l_1)}{2\pi}$

Multiply both sides by $4$ and substitute $l_2 = 20 - l_1$:

$l_1 = \frac{6(20 - l_1)}{\pi} = \frac{6l_2}{\pi}$

Now, rearrange to find the ratio $(\pi l_1) : l_2$:

$\pi l_1 = 6l_2$

$\frac{\pi l_1}{l_2} = 6$

Therefore, the ratio $(\pi l_1) : l_2$ is $6 : 1$.

7. Verify it's a Minimum (Second Derivative Test)

Calculate the second derivative, $\frac{d^2S}{dl_1^2}$:

$\frac{d^2S}{dl_1^2} = \frac{d}{dl_1}\left(\frac{l_1}{4} - \frac{3(20 - l_1)}{2\pi}\right) = \frac{1}{4} + \frac{3}{2\pi}$

Since $\frac{d^2S}{dl_1^2} > 0$, the critical point corresponds to a minimum.

Common Mistakes & Tips

• Chain Rule: Remember to apply the chain rule when differentiating $(20 - l_1)^2$. The derivative of $(20 - l_1)$ with respect to $l_1$ is $-1$.
• Algebraic Manipulation: Be careful with algebraic manipulations, especially when dealing with fractions and $\pi$.
• Second Derivative Test: Always confirm that the critical point you find corresponds to a minimum (or maximum) using the second derivative test.

Summary

This problem demonstrates a classic application of derivatives in optimization. We formulated the areas of the square and circle in terms of the lengths of wire used, created an objective function to minimize, used the constraint to reduce the problem to a single variable, and then found the critical point using differentiation. The second derivative test confirmed that the critical point corresponds to a minimum. The ratio $(\pi l_1) : l_2$ is $6 : 1$.

Final Answer The final answer is \boxed{6:1}, which corresponds to option (A).