If you have 36 feet of fencingwhat are the areas of the different rectangles you could enclose with the fencing?

Answer 1

To determine the areas of different rectangles that can be enclosed with 36 feet of Fencing, we can express the perimeter as ( P = 2(l + w) = 36 ), where ( l ) is the length and ( w ) is the width. This simplifies to ( l + w = 18 ). The area ( A ) of the rectangle can be expressed as ( A = l \times w ). By substituting ( w = 18 - l ) into the area formula, we find that the area varies depending on the values of ( l ) and ( w ), reaching a maximum area of 81 square feet when ( l = w = 9 ) (a square). Other combinations of length and width will yield varying areas, all less than or equal to 81 square feet.