008 Review Problem - Pint tin can

$V = 1 ~ \text{pint} \times \dfrac{1 ~ \text{gal.}}{8 ~ \text{pints}} \times \dfrac{231 ~ \text{in.}^3}{1 ~ \text{gal.}}$

$V = 28.875 ~ \text{in.}^3$
 

$V = \frac{1}{4}\pi d^2 h$

$28.875 = \frac{1}{4}\pi d^2 (4)$

$d^2 = 28.875/\pi$

$d = 3.0317 ~ \text{in.}$           answer
 

Total surface area of 1 can
$A_1 = 2A_b + A_L$

$A_1 = 2(\frac{1}{4}\pi d^2) + (\pi d)h$

$A_1 = \frac{1}{2}\pi (28.875/\pi) + \pi(3.0317)(4)$

$A_1 = 52.534\,951 ~ \text{in.}^2$
 

For 10,000 units
$A = 10\,000A_1$

$A = 10\,000(52.534\,951)$

$A = 525\,349.51 ~ \text{in.}^2 \left( \dfrac{1 ~ \text{ft.}}{12 ~ \text{in.}} \right)^2$

$A = 3648.26 ~\text{ft.}^2$           answer