Answer:
3.5%
Step-by-step explanation:
The volume of a cylinder = [tex]\pi r^2h[/tex]
r = radius of cylinder,
h = height of cylinder
For the non-optimal can,
r = 2.75/2 = 1.375
h = 5.0
[tex]V = \pi(1.375^2)\times 5.0 = 9.453125\pi[/tex]
For the optimal can,
d/h = 1,
d = h
2r = h
r = h/2
[tex]V = \pi\left(\dfrac{h}{2}\right)^2\times h = \pi\left(\dfrac{h^3}{4}\right)[/tex]
They have the same volume.
[tex]\pi\dfrac{h^3}{4} = 9.453125\pi[/tex]
[tex]h^3 = 37.8125[/tex]
[tex]h=3.36[/tex] (This is the height of the optimal can)
[tex]r = \dfrac{3.36}{2} = 1.68[/tex] (This is the radius of the optimal can)
The area of a cylinder is
[tex]A = 2\pi r(r+h)[/tex]
For the non-optimal can,
[tex]A = 2\pi\times\dfrac{2.75}{2}\left(\dfrac{2.75}{2}+5.0\right) = 17.53125\pi[/tex]
For the optimal can,
[tex]A = 2\pi\times1.68\left(1.68+3.36\right) = 16.9344\pi[/tex]
Amount of aluminum saved, as a percentage of the amount used to make the optimal cans = [tex]\dfrac{17.53125\pi - 16.9344\pi}{16.9344\pi}\times 100\% = 3.5\%[/tex]
