 # A rectangular playground is to be fenced off and divided in two by another fence parallel to one side of the playground. 400 feet of fencing is used. Find the dimensions of the playground that maximize the total enclosed area. What is the maximum area? This question aims to understand the geometry of the rectangle and how to calculate its area and parameter. Furthermore, it explains the quadratic equations and the quadratic formula. A rectangle has four sides. All the internal angles in the rectangle are equal to $90 \space degree$. Two sides produce a right angle when they meet at any corner. In a rectangle, only the opposite sides are equal in length, making it different from a square. To calculate the area of the rectangle, you have to multiply the length and the width. Area $A$ is $L$ times $W$. $Area=L \times W$ For calculating the perimeter, you simply add all the sides of the rectangle. there are 4 sides and the opposite sides are equal in length. The formula is given as: $Perimeter = 2(L+W)$ The quadratic formula enables solving any quadratic equation. First, we convert the equation in the form $ax²+bx+c=0$, where a, b, and c are coefficients. Then, we plug these coefficients in the formula: $\dfrac{-b \pm \sqrt{4ac}}{2a}$ and find the unknown $x$.

A rectangular playground is to be fenced off and split into two by another fence parallel to one side. Total fencing is $400$ feet. That means the 4 four sides of the rectangle $(2L+2W)$ and the one fence parallel to the one side $(W)$, a total of $5$, $(2L+3W)$ fences add up to $400$ the total fencing. The following equation represents the scenario. $2L+3W=400$ $3W=400-2L$ $W= \dfrac{400-2L}{3}$ $W= \dfrac{400}{3}- \dfrac{2L}{3}$ Area of the rectangle (in terms of $L$) $Area= \space W \times L$ Inserting $W$ and calculating Area in terms of $L$. $Area=W \left(\dfrac{400}{3} – \dfrac{2L}{3} \right)$ $=\left(\dfrac{400}{3}- \dfrac{2L}{3} \right)$ $=L \left(\dfrac{400}{3} – \dfrac{2L}{3} \right)$ $= -\dfrac{2}{3}L^{2}+ \dfrac{400}{3}L$ Now solving the quadratic equation using the formula: $L = \dfrac{-b \pm \sqrt{4ac}}{2a}$ Here: $a=-\dfrac{2}{3}$, $b = \dfrac{400}{3}$ and $c = 0$ $= \dfrac{- \dfrac{400}{3} \pm \sqrt{4(\dfrac{2}{3}) (0)}}{- 2 (\dfrac{2}{3})}$ $L = \dfrac{- \dfrac{400}{3}}{- 2 (\dfrac{2}{3})}$ By solving the expression, $L$ comes out to be: $L = 100$ Plugging in the $L= 100$ in the Width $W$ equation: $W= \dfrac{400}{3} – \dfrac{2L}{3}$ $W= \dfrac{400}{3} – \dfrac{2(100)}{3}$ $W = 66.6$ After calculating the Length and Width, calculate the area of the rectangle: $Area = L\times W$ $Area = 100 \times 66.6$ $Area = 6666.6$

The dimensions of the rectangular playground that maximize the total enclosed area is $Length = 100 \space feet$ and $Width= 66.6 \space feet$. The maximum area is $6666.6 \space feet^2$.
The length and width of the rectangle shape are given as $17 cm$ and $13 cm$ respectively. Find the area and the parameter. Given the Length $L = 17$ and Width $W=13$ The area of the rectangle is given by: $Area = L\times W$ $Area = 17 \times13$ $Area = 221 cm^2$ The parameter of the rectangle is given by: $Parameter = 2(L + W)$ $= 2(17 + 13)$ $= 2(30)$ $Parameter = 60 cm$