Final answer:
To maximize the area of a rectangular deck with one side attached to a building and 40 meters of fencing available for the other three sides, the optimal dimensions are 20 meters by 10 meters, resulting in a maximum area of 200 square meters.
Step-by-step explanation:
To determine the dimensions of a rectangular deck using a building side as one of the four required sides and fencing for the other three, with a total of 40 meters of fencing available, we employ optimization techniques from calculus. Let the side against the building be the length (L), and the side perpendicular to the building be the width (W). The perimeter of the rectangle will be represented as P = 2W + L. Given that only three sides need fencing, we set P equal to 40 meters, so 2W + L = 40.
Now, we solve for one variable in terms of the other, L = 40 - 2W, and substitute in the area formula A = L * W. Therefore, A = W * (40 - 2W) which simplifies to A = 40W - 2W^2. To find the maximum area, we calculate the derivative of the area with respect to W, set it to zero, and solve for W. The first derivative is A' = 40 - 4W. Setting it to zero gives us 40 - 4W = 0, leading to W = 10. Thus, the width that maximizes the area is 10 meters, and the corresponding length using the perimeter equation is L = 40 - 2(10) = 20 meters. Therefore, the deck with maximum area will have dimensions 20 meters by 10 meters, and the maximum area will be 200 square meters.