You are going to make many cylindrical cans. The cans will hold different volume

Question

You are going to make many cylindrical cans. The cans will hold different volumes. But you'd like them all to be such that the amount of sheet metal used for the cans is as small as possible, subject to the can holding the specific volume. How do you choose the ratio of height to diameter of the can? Assume that the thickness of the wall, top, and bottom of the can is everywhere the same, and that you can ignore the material needed for example to join the top to the wall Put differently. you ask what ratio of height to diameter will minimize the area of a cylinder with a given volume? That ratio equals Hint: You can do this the Calc I way by eliminating height or diameter, but the easier way is to use a Lagrange Multiplier

Explanation / Answer

Lets find the ratio betwwen radius and height first -

The surface area S is the sum of the areas of 2 circles of
radius r and a rectangle with height h and width 2r: Thus,
S = 2pir2 + 2pirh
This is constrained by a volume of r2h = 0:25 f t3
; so that the Lagrangian is
L(r; h; ) = 2r2 + 2rh r2h 0:25

r2h
0:25
Setting Lr = 0 and Lh = 0 leads to
4r + 2h = (2rh); 2r =

r2

The ratio of the two equations is
4r + 2h
2r
=
(2rh)
(r2)
=)
2r + h
r
=
2h
r
Cross-multiplying yields 2r
2 + rh = 2rh, which in turn yields
rh = 2r
2
or h = 2r
since r cannot be 0. That is, all cans with minimal surface area
will have h = 2r; which means a height equal to the diameter. To
determine which such can satisÖes the constraint, we substitute to
obtain
r2
(2r) = 0:25; r3 =
3
r
0:25
2
which leads to r = 0:3414 feet, with h = 0:6818 feet. To see that
a minimum must occur, we notice that the constraint implies that
h = 0:25=

r2

; which leads to S as a function of r in the form
S = 2r2 +
0:5
r
2
Straightforward di§erentiation shows that S
00 (r) > 0 for all r > 0;
so that any extremum must be a minimum.

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