you decide to fence in a rectangular yard behind your house. one side of the yard will be next to your…

you decide to fence in a rectangular yard behind your house. one side of the yard will be next to your house, and you have 126 feet of fencing available. what is the maximum possible area for the yard (rounded to the nearest tenth of a square foot)?

you decide to fence in a rectangular yard behind your house. one side of the yard will be next to your house, and you have 126 feet of fencing available. what is the maximum possible area for the yard (rounded to the nearest tenth of a square foot)?

Answer

Explanation:

Step1: Define variables

Let the length parallel to the house be (x) and the width perpendicular to the house be (y). The fencing constraint gives (x + 2y=126), so (x = 126 - 2y). The area (A=xy=(126 - 2y)y=- 2y^{2}+126y).

Step2: Find the vertex of the quadratic function

For a quadratic function (A(y)=ay^{2}+by + c) ((a=-2), (b = 126)), the (y) - coordinate of the vertex is (y=-\frac{b}{2a}). Substituting (a=-2) and (b = 126), we get (y=-\frac{126}{2\times(-2)}=\frac{126}{4}=31.5).

Step3: Find the value of (x)

Substitute (y = 31.5) into (x=126 - 2y). Then (x=126-2\times31.5=126 - 63 = 63).

Step4: Calculate the area

The area (A=xy). Substitute (x = 63) and (y = 31.5), so (A=63\times31.5 = 1984.5).

Answer:

(1984.5) square feet