Mathematics, 10.07.2019 02:30 donmak3833
How would i do part a. i showed my work but i'm slightly off
Answers: 1
![\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}](/tpl/images/0071/6491/b92a0.png)
![\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}](/tpl/images/0071/6491/c98ee.png)
![\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](/tpl/images/0071/6491/b5583.png)
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How would i do part a. i showed my work but i'm slightly off...
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