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?

Answer 1

52.1 yard²

Step-by-step explanation:

Let us assume the length of the center line is a yards and the other side is b yards. Since it is two adjacent identical rectangle, from the diagram attached, the perimeter of the rectangle is:

3a + 4b = 50 yards

Making a the subject of the formula:

`3a+4b=50\\a=(50-4b)/(3)`

The area of one rectangle is a× b, therefore for the two rectangles, the area is given as:

`A=2(a*b)\\Substituting:\\A=2((50-4b)/(3) )(b)\\A=(100b-8b^2)/(3)`

At maximum area, dA/db = 0

`A=(100b-8b^2)/(3)\\\\(dA)/(db)= (100)/(3)-(16b)/(3) \\at\ maximum\ area\ dA/db=0\\0=(100)/(3)-(16b)/(3) \\(100)/(3)=(16)/(3)b\\16b=100\\b= 100/16=6.25\ yard \\`

Substituting b to find a

`a=(50-4b)/(3)\\a=(50-4(6.25))/(3)=25/3=8.33`

The maximum area of one of the rectangle = ab = 8.33 × 6.25 = 52.1 yard²

The maximum area of the two adjacent identical rectangle = 2ab = 104.2 yard²