Sienna has 800 yards of fencing to enclose a rectangular area. Find the dimensions of the rectangle that maximize the enclosed area. What is the maximum area? A rectangle that maximizes the enclosed area has a length of \( \square \) yards and a width of \( \square \) yards. 0 of 2

To determine the dimensions of the rectangle that maximize the enclosed area with 800 yards of fencing, follow these steps: ### 1. Understanding the Problem - **Perimeter (P)**: The total length of fencing available is 800 yards. - **Rectangle Dimensions**: Let’s denote the length as \( L \) and the width as \( W \). ### 2. Relationship Between Length and Width The perimeter of a rectangle is given by: \[ P = 2L + 2W \] Given \( P = 800 \) yards: \[ 2L + 2W = 800 \] Simplifying: \[ L + W = 400 \quad \Rightarrow \quad W = 400 - L \] ### 3. Expressing Area in Terms of One Variable The area (\( A \)) of the rectangle is: \[ A = L \times W \] Substituting \( W \): \[ A = L \times (400 - L) = 400L - L^2 \] ### 4. Maximizing the Area To find the maximum area, we can take the derivative of \( A \) with respect to \( L \) and set it to zero: \[ \frac{dA}{dL} = 400 - 2L = 0 \] Solving for \( L \): \[ 2L = 400 \quad \Rightarrow \quad L = 200 \text{ yards} \] Substituting back to find \( W \): \[ W = 400 - 200 = 200 \text{ yards} \] ### 5. Maximum Area \[ A = 200 \times 200 = 40,000 \text{ square yards} \] ### **Final Answer** - **Length**: \( 200 \) yards - **Width**: \( 200 \) yards - **Maximum Area**: \( 40,\!000 \) square yards