A town decides to build a playground using one of the existing walls beside an elementary school. What is the maximum area they can enclose using only 100ft of donated fence?
Letting L be the length against the building,
100 = L + 2W
so, L = 100-2W
A = WL = W(100-2W)
= 100W - 2W^2
This has a max at W = 100/4 = 25
L = 50
A = 25*50 = 1250 ft^2
how to do input output tabol
To find the maximum area that can be enclosed using 100ft of donated fence, we need to determine the shape that would maximize the area. Let's consider different shapes and calculate their corresponding areas.
1. Rectangle:
Assuming the donated fence will be used for all four sides of the rectangle, we can express the perimeter of the rectangle in terms of the lengths of its sides: 2l + 2w = 100, where l represents the length and w represents the width. Rearranging the equation, we get l + w = 50. We want to find the maximum area, which is given by A = lw. To simplify, we can express w in terms of l: w = 50 - l. Substituting this into the area formula, we have A = l(50 - l) = 50l - l^2. To find the maximum area, we can take the derivative of A with respect to l and set it equal to zero: dA/dl = 50 - 2l = 0. Solving for l, we find l = 25. Substituting this value back into l + w = 50, we get 25 + w = 50, or w = 25. Therefore, the maximum area that can be enclosed using a rectangle is A = 25 * 25 = 625 square feet.
2. Square:
A square is a special case of a rectangle where all sides are equal. Using the same perimeter equation as before (l + w = 50), we can express both sides in terms of a single variable, l or w. Let's use l. Then, l = w = 50/2 = 25. The area of the square is A = l * l = 25 * 25 = 625 square feet, which is the same as the rectangle's maximum area.
Therefore, whether a rectangle or a square is used, the maximum area that can be enclosed using 100ft of donated fence is 625 square feet.