![\begin{gathered} a)1.353\text{ inches = radius max} \\ b)\text{Area}_s=105(in^2) \\ c)\text{Area}=93.5(in^2) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/kf22je1faszzzjhfh56cs582oq0j6fbyfq.png)
Step-by-step explanation
Step 1
a)
Let
![\text{length}=\text{ 11 inches}](https://img.qammunity.org/2023/formulas/mathematics/college/25wvb4z2uaxf9o46v8wpfbyq19r81m37bn.png)
the circle of the base will a arc of
![\text{perimeter}=\text{ 2 }\pi r](https://img.qammunity.org/2023/formulas/mathematics/college/mvo2amune101v98rpiqelwpbtbuex3uijd.png)
also, we know that perimeter of the circle equals the width of the paper, so
![\begin{gathered} 8(1)/(2)=2\text{ }\pi\text{ r} \\ 8.5=2\pi r \\ \text{divide both sides by 2}\pi \\ (8.5)/(2\pi)=(2\pi r)/(2\pi) \\ 1.3528\text{ inches = radius} \\ \text{rounded} \\ r=1.353 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/aga1ok0lewfdhh3hhfm53nzvofakqaywbx.png)
hence
the largest possible radius is 1.35 inches
Step 2
let
![\begin{gathered} \text{radius}=1.353\text{ in} \\ \text{height}=11\text{ in} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/fzqrej1q6d5kb041ajkbv1rt8mbarltibo.png)
the total surface area of a cylinder is given by
![\text{Area}_s=2\pi r(r+h)](https://img.qammunity.org/2023/formulas/mathematics/college/y5rcrw8vrxbzq66rak81rpe31jb8pqwj5r.png)
then, replace
![\begin{gathered} \text{Area}_s=2\pi r(r+h) \\ \text{Area}_s=2\pi(1.353\text{ in)(1.353 in+11 in)} \\ \text{Area}_s=2\pi(1.353\text{ in)(12.353 in)} \\ \text{Area}_s=105.01470(in^2) \\ \text{rounded} \\ \text{Area}_s=105(in^2) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/9jrnhc65ffkc1nxuo8o4hkcohzrjbde1vr.png)
Step 3
the area of the original paper
it is a rectangle, the area of a rectangle is given by:
![\text{Area}=\text{length }\cdot width](https://img.qammunity.org/2023/formulas/mathematics/college/hbp46hjxliwdvty1xyqj36otrb20scjos3.png)
Let
length= 11 inches
width=8.5 inches
replace
![\begin{gathered} \text{Area}=\text{length }\cdot width \\ \text{Area}=11\text{ in }\cdot8.5\text{ in} \\ \text{Area}=93.5(in^2) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/bat12cp37g4drwc00bz3s59zsotj93x3it.png)
I hope this helps you