Answer:
1250 [tex]ft^{2}[/tex]
Step-by-step explanation:
We are given that the playground is fenced on three sides of the playground and the four side has an existing wall.
Let the length of the rectangle be 'X' feet and width be 'Y' feet.
As the fencing is done using 100 feet of fence. We get the relation between the sides and the fence as, X + 2Y = 100.
As, X + 2Y = 100 → X = 100 - 2Y
Now, the area of the rectangle = XY = X × ( 100 - 2Y ).
i.e Area of the rectangle = [tex]-2Y^{2} +100Y[/tex].
The general form of a quadratic equation is [tex]y=ax^{2}+bx+c[/tex].
The maximum value of a quadratic equation is given by [tex]x=\frac{-b}{2a}[/tex].
Therefore, the greatest value of [tex]-2Y^{2} +100Y[/tex] is at Y = [tex]\frac{-100}{2 \times -2}[/tex] = [tex]\frac{-100}{-4}[/tex] = 25.
Thus, Y = 25 and X = 100 - 2Y → X = 100 - 2 × 25 → X = 50.
Hence, the area of the rectangle is XY = 50 × 25 = 1250 [tex]ft^{2}[/tex].
