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Optimization Problems

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3 Answers

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
 
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. 

Comments

I would hire James F. who commented on your problem.  
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!