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Optimization problem: what are the dimensions of the lightest open-top right circularcylindrical can that will hold a volume of 1000cm3?Compare the result with the result in example 2. ...?

Verified Answer :

So Volume of cylinder is pir^2h = 1,000 

Then lightest one means you have the smallest surface area. Which is one base and then the area of the surface. SA = pir^2 + 2pirh 

So now you have 2 equations, so: 

h = 1,000/(pi
r^2) 
So then SA = pir^2 + 2pir(1,000/(pir^2) = pir^2 + 2,000/r 

Derivative of SA is then 2pi
r -2,000/r^2. Set to 0 

2pir-2,000/r^2 =0 --> 2pir^3 = 2,000 --> r^3 = 1,000/pi --> r = 10/pi^(1/3) 

Now go back to the volume function: pir^2h =1,000 --> 1,000/(pi*100/pi^(2/3)) = h 
h = 10 / pi^(1/3)