Question

The “tin foil boat challenge” is a popular STEM
outreach activity for elementary school students (and adults too!).
It involves a contest to fold a sheet of tin foil into a shape
capable of supporting the most weight (usually quantified as the
most pennies) in a tub of water.

A popular design of tin foil boat is a square-bottomed vessel with
straight walls folded upwards to make a box-like boat (search
images of “tin foil boat” on the web). Starting with a square, flat
sheet of heavy-duty tin foil that measures 12 inches along each
side,

(a) determine the best way to fold the sheet (i.e. dimensions) in this fashion to maximize the volume, and therefore weight-bearing capacity.

(b) Then, calculate the number of pennies that the
boat could support before capsizing.

Answer #1

Consider that the tin foil vessel has a base of side a and a
height b. so, the volume of the tin foil is V= a^{2}b

Since the side of tin sheet is 12 inches, we can get that a+2b=12

b= 6 - a/2

therefore V=6a^{2} - a^{3}/2

To maximize volume, dV/da should be zero.

*dV**da**=12a-**3**2*
*a**2* =0

**a=12 x 2/3 = 8 inches.**

and **b= 2 inches.**

b) The volume of the above configuration is 8*8*2 =128 cubic inches.

The boat will displace same amount of water before sinking.

This is 2.098 liters of water (divide volume by 61.024)

The weight of same amount of water is 2.098 kilograms.

The weight of a penny is 2.5 grams ie: 2.5 x 10^{-3}
kilograms. Dividing the weight of water by this will give the
number of pennies as **839 pennies**

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