## Algebra 2 (1st Edition)

The surface area is: $2\pi rl+2(4/2\pi r^2)=2\pi rl+4\pi r^2$ The volume is: $r^2\pi l+2(4/3/2\pi r^3)=r^2\pi l+4/3\pi r^3$ The ratio is: $\frac{2\pi rl+4\pi r^2}{r^2\pi l+4/3\pi r^3}=\frac{2\pi r(l+2r)}{\pi r(rl+4/3r^2)}=\frac{2(l+2r)}{(rl+4/3r^2)}=\frac{6(l+2r)}{r(3l+4r)}$ Thus we proved what we had to.