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Rina8888 [55]
3 years ago
7

How would I do part A. I showed my work but I'm slightly off

Mathematics
1 answer:
ahrayia [7]3 years ago
5 0

\bf \stackrel{\textit{perimeter of the rectangle}}{P_{rect}=2x+y}~\hfill \stackrel{\textit{area of the rectangle}}{A_{rect}=xy}
\\\\\\
\stackrel{\textit{perimeter of the semi-circle, with }r=\frac{y}{2}}{P_{semic}=\cfrac{\pi y}{2}}~\hfill \stackrel{\textit{area of the semi-circle with }r=\frac{y}{2}}{A_{semic}=\cfrac{\pi y^2}{8}}
\\\\[-0.35em]
\rule{34em}{0.25pt}


\bf \stackrel{\textit{perimeter of the enclosure}}{60=2x+y+\cfrac{\pi y}{2}}\implies \stackrel{\textit{multiplying by 2}}{120=4x+2y+\pi y}
\\\\\\
120-2y-\pi y=4x\implies \cfrac{120-2y-\pi y}{4}=x
\\\\[-0.35em]
\rule{34em}{0.25pt}


\bf \stackrel{\textit{area of the enclosure}}{A=xy+\cfrac{\pi y^2}{8}}\implies A=\left( \cfrac{120-2y-\pi y}{4} \right)y+\cfrac{\pi y^2}{8}
\\\\\\
A=\cfrac{120y-2y^2-\pi y^2}{4}+\cfrac{\pi y^2}{8}\implies A=\cfrac{240y-4y^2-2\pi y^2+\pi y^2}{8}
\\\\\\
A=\cfrac{240y-4y^2-\pi y^2}{8}\implies A=\cfrac{240y}{8}-\cfrac{4y^2}{8}-\cfrac{\pi y^2}{8}
\\\\[-0.35em]
\rule{34em}{0.25pt}\\\\
~\hfill A=30y-\cfrac{1}{2}y^2-\cfrac{1}{8}\pi y^2~\hfill

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