Design a Better Drink Can
Author: Steve Arnold
Date: 15/10/2006

A standard can of drink holds 375 ml.
Most are around 6cm across the base, and 13 cm high.

But is this the BEST size?

Since the cost of materials is most closely related to the surface area of the can, we would like dimensions that give the greatest volume for the smallest surface area.

Drag the point shown to change the radius and observe the graph to see how the surface area (divided by 100!) changes! Try different volumes.

Try and find the algebraic function which matches this situation.

What assumptions have we made in building this model, and how might these be minimised?

Drag me!

Top View

Side View

Surface Area =

Target Volume

Design a Better Drink Can

A standard can of drink holds 375 ml.

Most are around 6cm across the base, 
and 13 cm high.

But is this the BEST size?

Investigate!

Thickness of
Top and Base

n = 1 (slider from 0 to 4, step 0.1)
v = 375 (slider from 300 to 500, step 5)
O = (20, 18)
a = Line(O, xAxis)
A = Point(a)
r = Segment(O, A)
E: (20.0 + r, 18.0 - r - v / (3.14 * r * r))
B: (20.0 + r, 18.0 - r)
c = Circle(O, A)
b = OrthogonalLine(A, xAxis)
d = OrthogonalLine(B, b)
C = Mirror(A, O)
e = OrthogonalLine(C, d)
D = Intersect(d, e)
f = OrthogonalLine(E, b)
F = Intersect(e, f)
P, d_1, h, e_1, f_1 = Polygon(D, B, E, F)
W: (r, 0.02 * (3.14 * n * r * r * r + v) / r)
G = (25.32075471698114, 19.794116313199208)
u = Vector(G, A)
H = (26, 19)
I = (41, 19)
J = (41, 4)
K = (26, 4)
Q, h_1, i, j, k = Polygon(H, I, J, K)
S: (20.0 + n / 4.0, 18.0)
p = Circle(S, r)
T6: 2.0 * n * 3.14 * r * r + 6.28 * r * h + ""
loc1 = Locus(W, A)
loc2 = Locus(W, A)
T10: v + ""