
Expert Answer
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: \[Area = L\times W\] \[Area = 100 \times 66.6\] \[Area = 6666.6 \]Numerical Answer
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 $6666.6 \space feet^2$.Example
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: \[Parameter = 2(L + W)\] \[= 2(17 + 13)\] \[= 2(30)\] \[Parameter = 60 cm\]Previous Question < > Next Question
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