Answer:
[tex]A_{r} \approx 515.044\,cm^{2}[/tex], [tex]s\approx 376.991\,cm[/tex]
Explanation:
El área total de la lamina es [tex]2400\,cm^{2}[/tex] (The total area of the sheet is [tex]2400\,cm^{2}[/tex]). La función asociada al área sobrante es (The function associated with the remaining area is):
[tex]A_{r} = 2400\,cm^{2} - 6\cdot \pi \cdot r^{2}[/tex]
El radio máximo está asociado con la longitud más corta de la lámina. Asúmase que [tex]r = 10\,cm[/tex] (The maximum radius is associated with the smallest length of the sheet. Let assume that [tex]r = 10\,cm[/tex]):
[tex]A_{r} = 2400\,cm^{2} - 6\cdot \pi \cdot (10\,cm)^{2}[/tex]
[tex]A_{r} \approx 515.044\,cm^{2}[/tex]
La longitud necesaria de hule de protección es igual a: (The required length of proctection rubber is equal to)
[tex]s = 12\cdot \pi \cdot (10\,cm)[/tex]
[tex]s\approx 376.991\,cm[/tex]
