A carpenter is building a rectangular room with a fixed perimeter of 360 ft . What dimensions would yield the maximur Length \( (\mathrm{ft})=90 \) Width \( (\mathrm{ft})=\square \)

To maximize the area of the rectangular room while maintaining a fixed perimeter of 360 ft, we can start by determining the width. The perimeter formula is \( P = 2L + 2W \). With a length of 90 ft, we can rearrange this to find the width: \[ 360 = 2(90) + 2W \] This simplifies to \( 360 = 180 + 2W \), leading to \( 2W = 180 \), so \( W = 90 \) ft. Interestingly, for fixed perimeters, a square configuration often provides maximum area. In this case, both the length and width being 90 ft forms a square shape, yielding an area of \( 8100 \, \text{ft}^2 \). It's efficient and widely applicable! To solidify your understanding, remember that when you change one dimension while keeping the perimeter constant, the other dimension must adjust accordingly, which is crucial in practical carpentry and design projects! Always double-check your calculations to avoid sizing mistakes.