Question

The ubiquitous 12oz aluminum cans used to distribute drinks in this country have a diameter of approximately 2.75 inches and a height of 5.0 inches. Estimate the amount of aluminum that could be saved if the cans were designed with the ratio d/h=1, while having the same volume as the standard 12oz can. Amount of aluminum saved, as a percentage of the amount used to make the optimal cans, equals ______

Answer 1

Answer:

3.5%

Step-by-step explanation:

The volume of a cylinder = $\pi r^2h$

r = radius of cylinder,

h = height of cylinder

For the non-optimal can,

r = 2.75/2 = 1.375

h = 5.0

$V = \pi(1.375^2)\times 5.0 = 9.453125\pi$

For the optimal can,

d/h = 1,

d = h

2r = h

r = h/2

$V = \pi\left(\dfrac{h}{2}\right)^2\times h = \pi\left(\dfrac{h^3}{4}\right)$

They have the same volume.

$\pi\dfrac{h^3}{4} = 9.453125\pi$

$h^3 = 37.8125$

$h=3.36$ (This is the height of the optimal can)

$r = \dfrac{3.36}{2} = 1.68$ (This is the radius of the optimal can)

The area of a cylinder is

$A = 2\pi r(r+h)$

For the non-optimal can,

$A = 2\pi\times\dfrac{2.75}{2}\left(\dfrac{2.75}{2}+5.0\right) = 17.53125\pi$

For the optimal can,

$A = 2\pi\times1.68\left(1.68+3.36\right) = 16.9344\pi$

Amount of aluminum saved, as a percentage of the amount used to make the optimal cans = $\dfrac{17.53125\pi - 16.9344\pi}{16.9344\pi}\times 100\% = 3.5\%$