Optimization problems: What are the optimal dimensions of a can of this volume?

GeekOfMath

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Hello!
I need help with a couple of optimization problems that I'm unsure of how to begin. Any advice or help is welcome and much appreciated.

Question 1:
Collect data on several cans with the same mass of product but different sizes and collate the information in a table (already have)
Product NameCan 1Can 2
Printed volume300g300g
Can Dimensions
Height
Diameter
Height – Approx 8cm
Diameter – Approx 6.9cm
Height – Approx 10.1cm
Diameter – Approx 6.1cm
Actual volume of each canV = πr2h
= π x (6.9/2)2 x 8
= π x 3.452 x 8
= π x 11.9 x 8
= 299.08cm3
V = πr2h
= π x (6.1/2)2 x 10.1
= π x 3.052 x 10.1
= π x 9.3 x 10.1
= 295.09cm3
The surface area of each canA = 2πrh + 2πr2
= 2π (6.9/2) x 8 + 2π x (6.9/2)2
= 2π x 3.45 x 8 + 2π x 3.452
= 248.2cm2
A = 2πrh + 2πr2
= 2π (6.1/2) x 10.1 + 2π (6.1/2)2
= 2π x 3.05 x 10.1 + 2π x 3.052
= 252cm2

A) Which can is the most economical? - i.e. the lowest ration of volume to surface area
B) What are the optimal dimensions of a can of this volume?

So I am a little confused on how to start answering these questions. It would be excellent if someone could me a pointer on where to start. I get the feeling A needs to a function relating surface area to volume ratio and then differentiated and then equaled to 0 to get the minimum value but still unsure of how to go about doing this. Thanks in advance!
 
The first problem asks, "Which of these cans has the lowest ratio of volume to surface area". Since you have already calculated both volume and surface area, just divide! Since a can is designed to hold a specific volume but you have to pay more for greater surface area (more metal), the "optimum shape" would be the largest ratio of volume to surface area.

Two cans is probably not enough to decide a general rule for "optimum shape".
 
The first problem asks, "Which of these cans has the lowest ratio of volume to surface area". Since you have already calculated both volume and surface area, just divide! Since a can is designed to hold a specific volume but you have to pay more for greater surface area (more metal), the "optimum shape" would be the largest ratio of volume to surface area.

Two cans is probably not enough to decide a general rule for "optimum shape".

Ok, I think I get it. Do I divide to get a smaller volume to surface area ratio? Or do you mean something else? Also we have been studying optimization so I'm a little unsure how to incorporate what we have learnt into the answers and working out. Thanks for replying!
 
Do I divide to get a smaller volume to surface area ratio?
For each can, you divide in order to see what the ratio's value is (in decimal form). Compare these values; the smaller of the two shows you which can is the solution, for part (a).
 
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