Problem 185 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

A wire of length 2 units is cur into two parts which are bent respectively to form a square of side \(=x\) units and a circle of radius \(=\mathrm{r}\) units. If the sum of the areas of the square and the circle so formed is minimum, then: (A) \(2 x=r\) (B) \(2 x=(\pi+4) r\) (C) \((4-\pi) x=\pi r\) (D) \(x=2 r\)

Step-by-Step Solution

Answer

The condition for minimum area is (B) \(2 x=(\pi+4) r\).

Step 1: Express the perimeter conditions

Let's divide the wire of length 2 units into two parts. One part will have a length which forms the perimeter of the square, and the other the circumference of the circle. Say that the part forming the square has a length of 4x, because a square side of length x has a perimeter of 4x. The part forming the circle will then have a circumference of \(2\pi r\), where r is the radius of the circle. Thus, we have the equation: \[ 4x + 2\pi r = 2 \] solving for x gives: \[ x = \frac{2 - 2\pi r}{4} \]

Step 2: Derive the areas in terms of variables

The area of the square is \(x^2\) and the area of the circle is \(\pi r^2\). The objective is to minimize the sum of these two areas.

Step 3: Express the objective function

Substitute \(x\) from the equation in Step 1 into the area of the square: \[ x = \frac{2 - 2\pi r}{4} \Rightarrow x^2 = \left(\frac{2 - 2\pi r}{4}\right)^2 \] Thus, the total area to minimize (sum of the square and circle areas) becomes: \[ \text{Area} = \left(\frac{2 - 2\pi r}{4}\right)^2 + \pi r^2 \] Simplify this equation to: \[ \frac{(2 - 2\pi r)^2}{16} + \pi r^2 \]

Step 4: Minimize the objective function

Use calculus to minimize the obtained area function with respect to \(r\). Differentiate the area function with respect to \(r\) and set the derivative to zero to find the critical points. Solve: \[ \frac{d}{dr}\left(\frac{(2 - 2\pi r)^2}{16} + \pi r^2 \right) = 0 \] Simplifying the derivative and setting it to zero will result in the condition for minimizing areas.

Step 5: Solve the derivative equation

Upon solving the derivative equation, you will find that a critical point occurs at \(2 x = (\pi + 4) r\), indicating that this is the condition for the minimum area.

Key Concepts

Perimeter Conditions

When dealing with optimization problems involving shapes, one essential step is defining how a total length is divided among the shapes. In this exercise, a wire of 2 units is split between a square and a circle.
The perimeter of each shape must add up to the total length of the wire. For the square, with side length of \(x\), the perimeter is \(4x\). The circle has a circumference of \(2\pi r\). Combining these, we get:
\[4x + 2\pi r = 2\]Here, the constraint equation ensures that our lengths do not exceed the total wire length. Notice that solving this equation for one variable allows substitution in later steps, which is crucial for forming the objective function.

Objective Function

The primary goal in optimization problems is usually to maximize or minimize a function, known as the objective function. In this scenario, we seek to minimize the total area formed by the square and the circle using the wire.
The area of the square is \(x^2\) and the area of the circle is \(\pi r^2\). The objective function, representing the total area, is given by:
\[\text{Total Area} = x^2 + \pi r^2\]By substituting \(x\) from the perimeter equation, the objective function in terms of \(r\) is:
\[\left(\frac{2 - 2\pi r}{4}\right)^2 + \pi r^2\]This transformation is the key to simplifying the problem so it can be solved using calculus methods.

Calculus Method

Using calculus to find a minimum involves taking the derivative of an equation. Here, you need the derivative with respect to \(r\) to find at which points the total area will be minimized.
**Steps in the Calculus Method:**

• Differentiate the objective function with respect to \(r\).
• Set the derivative equal to zero. This is because the slope of the function at its minimum (or maximum) is zero.

After differentiating the given function:
\[\frac{d}{dr}\left(\frac{(2 - 2\pi r)^2}{16} + \pi r^2\right)\]you determine when the sum of the areas is at its minimum by finding the values of \(r\) fitting this condition. This procedure helps identify critical points.

Critical Points

Critical points are where the derivative equals zero, indicating potential minima or maxima. Once identified, further analysis may determine whether these points are where the area truly is minimized.
For this exercise, solving the derivative condition leads to:
\[ 2x = (\pi + 4)r \]This critical point states the relationship at which the areas are minimized under the given perimeter constraint. These are potential conditions for minimizing the area. Verifying whether this critical point yields a minimum involves second derivative tests or analyzing the problem constraints.