A physical education teacher is planning to outline two adjacent identical rectangular areas for a new game that students will be learning. If the boundaries, including the center line, of this "court" must be marked with a single 50 -yard roll of tape, what is the maximum area of one of the smaller rectangular spaces, rounded to the nearest tenth of a square yard?

Question

A physical education teacher is planning to outline two adjacent identical rectangular areas for a new game that students will be learning. If the boundaries, including the center line, of this "court" must be marked with a single 50   -yard roll of tape, what is the maximum area of one of the smaller rectangular spaces, rounded to the nearest tenth of a square yard?

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Calculate Total Perimeter: First, we need to determine the total length of tape used to outline the two identical rectangular areas including the center line. Since there are two rectangles sharing one side (the center line), the total perimeter that the tape needs to cover is the perimeter of one rectangle plus one additional length (the center line).Set Up Equation: Let's denote the length of the rectangle as L and the width as W. The perimeter of one rectangle is 2L + 2W. Since there are two rectangles sharing one side, the total length of tape used will be 2L + 2W + W (the additional W is the center line).Maximize Area: Now we can set up the equation for the total length of tape used: 2L + 3W = 50 yards (since we have a 50-yard roll of tape).Assume Equal Length and Width: To maximize the area of one of the rectangles, we need to use the entire length of tape. We also know that for a given perimeter, a rectangle's area is maximized when the rectangle is a square. However, since we have two rectangles and an extra length (the center line), it won't be a perfect square. But we can still use this principle to get close to the maximum area by trying to keep L and W as close to each other as possible.Solve for Width: Let's assume that the length and width are equal, which would make the shape a square. This would mean L = W. However, we know that we have an extra W to account for the center line. So let's set 2L + 3W equal to 50 and substitute L with W to see if we can find a solution: 2W + 3W = 50, which simplifies to 5W = 50.Check Solution: Solving for W, we get W = 50 / 5 = 10 yards. This would mean that L = W = 10 yards if it were a square, but we need to check if this fits our original equation 2L + 3W = 50 yards.Adjust Length and Width: Plugging L = 10 and W = 10 into the equation: 2(10) + 3(10) = 20 + 30 = 50 yards. This confirms that our assumption fits within the length of tape we have.However, we made an error in our assumption. Since we have an extra W (the center line), making L and W equal would not use the tape optimally. We need to adjust either L or W so that 2L + 3W still equals 50, but the area (L × W) is maximized. We can do this by decreasing W slightly and increasing L by the same amount to keep the total length of tape used the same.

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