Question

A wire of length 36 meters is bent to form a rectangle. Find its dimensions if the area of the rectangle is maximum.

Hint:

Maximum area for a rectangle with fixed perimeter is a square; use calculus to confirm.

Answer 1

Perimeter: \( 2(l + w) = 36 \Rightarrow l + w = 18 \Rightarrow w = 18 - l \). 
Area: \( A = l \cdot w = l (18 - l) = 18l - l^2 \). 
Maximize: \( A(l) = 18l - l^2 \). 
\[ \frac{dA}{dl} = 18 - 2l = 0 \Rightarrow l = 9. \] \[ \frac{d^2 A}{dl^2} = -2 < 0, \text{ maximum at } l = 9. \] \[ w = 18 - 9 = 9. \] Area: \( 9 \cdot 9 = 81 \, \text{m}^2 \). 
Answer: Dimensions: \( 9 \, \text{m} \times 9 \, \text{m} \).