# 008 Review Problem - Pint tin can

**Problem 8**

Find the diameter of the base of the pint tin can whose height is 4 in. (231 cu. in. = 1 gal.); also, find the number of square feet of tin required to manufacture 10,000 such cans with tops. (Neglect the waste due to seams, etc.)

**Solution 8**

$V = 28.875 ~ \text{in.}^3$

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

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

$d^2 = 28.875/\pi$

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

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