A soda can is hold 12 fluid ounces. Suppose that the bottom and the top are twice thick as the sides. Find the dimensions of the can that minimize the amount of the material used(Hint: instead of minimizing the surface area minimize the cost)
The dimensions of the can that minimize the amount of material used are a radius of 2 inches and a height of 6 inches.
AAAaannndd the bot gets it wrong yet again!
12 fl oz = 21.65 in^3
so since πr^2 h = 21.65, h = 21.65/(πr^2)
The cost is
A = 2*2πr^2 + 2πrh = 4πr^2 + 2πr*21.65/(πr^2) = 4πr^2 + 43.3/r
dA/dr = 8πr - 43.3/r^2
so A is minimum of 54.18, when r = 1.199 and h = 4.79