Final answer:
To maximize the area with 40m of plastic stripping, the court dimensions should be 10m by 20m, giving the maximum area of 200 square meters.
Step-by-step explanation:
To maximize the area of a rectangular handball court that is to be surrounded by 40 meters of plastic stripping on three sides, we must find the dimensions that give the largest area. We start with the perimeter equation for three sides of a rectangle because one side is a brick wall and does not need plastic stripping:
P = 2w + l = 40 meters, where w is the width and l is the length of the court.
To find the maximum area, we express the length in terms of the width using the perimeter equation: l = 40 - 2w.
Now we can write the area (A) as a function of width: A = w × l = w × (40 - 2w).
The area as a function of width is a quadratic equation: A(w) = -2w^2 + 40w.
To find the maximum area, we take the derivative of A with respect to w and set it to zero to find the critical points.
dA/dw = -4w + 40 = 0.
By solving for w, we find that w = 10 meters. Substituting w back into the equation for l gives us l = 40 - 2×(10) = 20 meters.
Therefore, the dimensions of the court that will give the maximum area are 10 meters by 20 meters.
The maximum area of the court is A = 10m × 20m = 200 square meters.