To find the dimensions of a rectangle of greatest area with a fixed perimeter \( p \), we can use some fundamental concepts from geometry and calculus. Let's break this down step-by-step. ### Step 1: Understanding the Problem We need to maximize the area \( A \) of a rectangle given a fixed perimeter \( p \). The perimeter \( P \) of a rectangle with length \( l \) and width \( w \) is given by the formula: \[ P = 2l + 2w \] Since the perimeter is fixed, we can express one variable in terms of the other. Rearranging the perimeter formula gives: \[ l + w = \frac{p}{2} \] From this, we can express \( w \) in terms of \( l \): \[ w = \frac{p}{2} - l \] ### Step 2: Expressing the Area The area \( A \) of the rectangle is given by: \[ A = l \times w \] Substituting for \( w \) from the previous step, we have: \[ A = l \left( \frac{p}{2} - l \right) \] This simplifies to: \[ A = \frac{p}{2}l - l^2 \] ### Step 3: Maximizing the Area To find the maximum area, we need to take the derivative of \( A \) with respect to \( l \) and set it to zero: \[ \frac{dA}{dl} = \frac{p}{2} - 2l \] Setting the derivative equal to zero gives: \[ \frac{p}{2} - 2l = 0 \] Solving for \( l \): \[ 2l = \frac{p}{2} \implies l = \frac{p}{4} \] ### Step 4: Finding the Width Now that we have \( l \), we can find \( w \) using the relationship we derived earlier: \[ w = \frac{p}{2} - l = \frac{p}{2} - \frac{p}{4} = \frac{p}{4} \] ### Step 5: Conclusion Thus, both the length and width of the rectangle that maximizes the area are \( \frac{p}{4} \). This means the rectangle is actually a square with each side measuring \( \frac{p}{4} \). ### Answer The correct option is **D. square of sides \( \frac{p}{4} \)**. ### Explanation of Other Options - **Option A: square of sides \( p \)** - This is incorrect because if each side were \( p \), the perimeter would be \( 4p \), not \( p \). - **Option B: square of sides \( 2p \)** - This is also incorrect for the same reason; the perimeter would be \( 8p \). - **Option C: square of sides \( \frac{p}{2} \)** - This would give a perimeter of \( 4 \times \frac{p}{2} = 2p\), which is not equal to \( p \). ### Common Pitfalls - Confusing the perimeter with the area. Remember that the perimeter is a linear measure while the area is a measure of space. - Not recognizing that the maximum area for a given perimeter occurs when the rectangle is a square. ### Revision Summary - The area of a rectangle is maximized when it is a square. - For a fixed perimeter \( p \), the dimensions of the square are \( \frac{p}{4} \) for each side. - The relationship between perimeter and dimensions is crucial: \( P = 2l + 2w \). - Always check the dimensions against the fixed perimeter to ensure correctness.