A cylindrical can is to have a volume of 1L.
a) Find the height and radius of the can that will minimize the surface area
1 liter = 1000 cm^{3}
Let x = the radius of the top and bottom of the can
Let y = the height of the can
The volume of the can is: V = (pi)x^{2}y = 1000 cm^{3}. So y = 1000/(pi)x^{2}
The surface area of the can is: S = 2(pi)x^{2} + 2(pi)xy
Substituting 1000/(pi)x^{2} for y, the surface area is:
S = 2(pi)x^{2} + 2(pi)x*1000/(pi)x^{2} = 2(pi)x^{2} + 2000x^{-1}
Take the derivative of S wrt x:
S' = dS/dx = 4(pi)x - 2000x^{-2}
S' = 0 = 4(pi)x - 2000x^{-2}
2000x^{-2} = 4(pi)x
2000/4(pi) = x^{3}
x = (2000/4(pi))^{1/3} = 10(1/2(pi))^{1/3} ≈ 5.42 cm
y = 1000/(pi)x^{2} = 10.84 cm
Check:
V = (pi)(5.42)^{2}(10.84) = 1000
b) What is the ratio of the height to the diameter?
The ratio of the height (y) to the diameter (2x) = 1
Do pop cans have a similar ratio? If not, why? You're on your own for this question! ;-)
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