Optimization Problems

Question

Liz Energy drink is to be sold in 472 mL cylindrical cans. Since it is a new product the package designers want the cans to be more visible when displayed with competing products so they want the height of the can to be no less than 12 cm. Determine the dimensions of the can that will minimize the surface area.

6/20/2014 | Shreya from Pleasantville, IA

Answer 1

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Optimization Problems

Liz Energy drink is to be sold in 472 mL cylindrical cans. Since it is a new product the package designers want the cans to be more visible when displayed with competing products so they want the height of the can to be no less than 12 cm. Determine the dimensions of the can that will minimize the surface area.

6/20/2014 | Shreya from Pleasantville, IA

Answer 2

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Optimization Problems

Answer 3

Solver Options
Max Time Unlimited, Iterations Unlimited, Precision 0.000001, Use Automatic Scaling
Convergence 0.0001, Population Size 100, Random Seed 0, Derivatives Forward, Require Bounds
Max Subproblems Unlimited, Max Integer Sols Unlimited, Integer Tolerance 1%, Assume NonNegative

Objective Cell (Min)
Cell Name Original Value Final Value
$D$3 area 345.4542674 345.4542674

Variable Cells
Cell Name Original Value Final Value Integer
$D$4 r 3.53838381 3.53838381 Contin
$D$5 h 12 12 Contin

The answer is Radius of the base = 3.54 cm

Height 12 cm and min surface area = 345.45

Volume 472 cm3

Constraints
Cell Name Cell Value Formula Status Slack
$D$6 volume 472.0000153 $D$6=472 Binding 0
$D$5 h 12 $D$5>=12 Binding 0

Answer 4

http://www.scribd.com/doc/15508287/Materials-Engineering-Science-Processing-and-Design.

I would advise looking at sites specifically linked to the manufacturing of "cylindrical cans" industries. This site might have a "Tutor" of such concepts, those related to complex mathematics and cans, thereof. I obviously do not have those qualifications, even with Two Master's degrees. Peace. Samuel.

6/20/2014 | Samuel T.

Answer 5

First, we want formulas for the 2 values of interest: Volume and Surface Area

Volume = (pi)(r^2)h

Surface Area = 2(pi)r^2 + 2(pi)rh

We want to minimize SA, but it has two variables, so we need to use the Volume equation to substitute.

Volume = 472 = (pi)(r^2)h --> r = sqrt[472/(pi*h)]

Now we can plug this into the SA formula

SA = 2(pi)[472/(pi*h)] + 2(pi)sqrt[472/(pi*h)]h

Now we have SA in terms of only one variable (h). All that's left is to differential SA with respect to h, set it equal to 0, and solve for h. Plug a value smaller than h and larger than h in to double check it is in fact a minimum.

Finally, use the h you found and the formula r = sqrt[472/(pi*h)] to solve for r.

Give this a shot and let me know if you have any more questions!

Optimization Problems

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Optimization Problems

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