Key Concepts and Formulas

• Optimization using Calculus: Finding the maximum or minimum value of a function subject to constraints using derivatives.
• Area of a Square: $A = x^2$, where $x$ is the side length.
• Circumference and Area of a Circle: $C = 2\pi r$ and $A = \pi r^2$, where $r$ is the radius.

Step-by-Step Solution

Step 1: Define Variables and Formulate the Constraint Equation

We define the variables:

• $x$: side length of the square.
• $r$: radius of the circle.

The total length of the wire is 2 units. This provides our constraint. The wire is cut into two parts, one forming the square and the other the circle. Therefore:

Perimeter of square + Circumference of circle = Total length of wire

$4x + 2\pi r = 2$

Dividing the equation by 2 simplifies it:

$2x + \pi r = 1 \quad \text{(Equation 1)}$

This equation links $x$ and $r$.

Step 2: Formulate the Objective Function (Sum of Areas)

We want to minimize the total area, which is the sum of the area of the square and the area of the circle.

• Area of square: $A_s = x^2$
• Area of circle: $A_c = \pi r^2$

The total area S is:

$S = A_s + A_c = x^2 + \pi r^2 \quad \text{(Equation 2)}$

This is the objective function we aim to minimize.

Step 3: Express the Objective Function in a Single Variable

To use calculus for optimization, we need to express the objective function S in terms of a single variable. We can use Equation 1 to express x in terms of r (or vice versa). Let's express x in terms of r:

From Equation 1: $2x + \pi r = 1$ $2x = 1 - \pi r$ $x = \frac{1 - \pi r}{2}$

Now, substitute this expression for x into Equation 2:

$S(r) = \left(\frac{1 - \pi r}{2}\right)^2 + \pi r^2$

Expanding and simplifying:

$S(r) = \frac{1 - 2\pi r + \pi^2 r^2}{4} + \pi r^2$ $S(r) = \frac{1}{4} - \frac{\pi}{2}r + \frac{\pi^2}{4}r^2 + \pi r^2$ $S(r) = \left(\frac{\pi^2}{4} + \pi\right)r^2 - \frac{\pi}{2}r + \frac{1}{4}$

This is S as a function of r only.

Step 4: Differentiate and Find Critical Points

To find the minimum of S(r), we differentiate with respect to r and set the derivative to zero:

$\frac{dS}{dr} = 2\left(\frac{\pi^2}{4} + \pi\right)r - \frac{\pi}{2}$ $\frac{dS}{dr} = \left(\frac{\pi^2}{2} + 2\pi\right)r - \frac{\pi}{2}$

Set $\frac{dS}{dr} = 0$:

$\left(\frac{\pi^2}{2} + 2\pi\right)r = \frac{\pi}{2}$

Step 5: Solve for r

Solve for r:

$\left(\frac{\pi^2 + 4\pi}{2}\right)r = \frac{\pi}{2}$ $(\pi^2 + 4\pi)r = \pi$ $\pi(\pi + 4)r = \pi$

Since $\pi \neq 0$, divide both sides by $\pi$:

$(\pi + 4)r = 1$ $r = \frac{1}{\pi + 4}$

Step 6: Solve for x and Find the Relationship Between x and r

Substitute the value of r back into the equation for x:

$x = \frac{1 - \pi r}{2} = \frac{1 - \pi \left(\frac{1}{\pi + 4}\right)}{2}$ $x = \frac{1 - \frac{\pi}{\pi + 4}}{2} = \frac{\frac{\pi + 4 - \pi}{\pi + 4}}{2} = \frac{\frac{4}{\pi + 4}}{2} = \frac{4}{2(\pi + 4)}$ $x = \frac{2}{\pi + 4}$

Now we have:

$x = \frac{2}{\pi + 4}$ $r = \frac{1}{\pi + 4}$

Therefore, the relationship between x and r is:

$x = 2r$

Common Mistakes & Tips

• Algebraic Manipulation: Carefully perform algebraic manipulations to avoid errors when simplifying equations and substituting variables.
• Chain Rule: Remember the chain rule when differentiating composite functions.
• Second Derivative Test: Although not strictly required here, remember to use the second derivative test to confirm whether a critical point is a minimum or a maximum.

Summary

We minimized the sum of the areas of a square and a circle formed from a wire of length 2. We first established a constraint equation relating the side of the square and the radius of the circle. Then, we formulated the total area as a function of a single variable, differentiated it, and set the derivative equal to zero to find the critical point. Finally, we found the relationship between x and r, which is $x = 2r$. This corresponds to option (A).

Final Answer The final answer is \boxed{x=2r}, which corresponds to option (A).