# Determine the dimensions of the cheapest possible container. In the following statement. A cylindrical container, with a volume of 4000`cm^3`, is being constructed to hold candies. The cost of the base and lid is \$0.005/`cm^2`, and the cost of the side walls is \$0.0025/ `cm^2` ----------------------------------------------------------------- Answer is r=6.8 cm and h=27.5 cm. Why?

You need to evaluate the total surface area of a cylinder such that:

`TSA = 2*pi*r*h`  (area of side surface) + `2pir^2` (area of the base and the lid)

The problem provides the information that the volume of cylinder is of `4000 cm^3` .

You need to write the formula...

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You need to evaluate the total surface area of a cylinder such that:

`TSA = 2*pi*r*h`  (area of side surface) + `2pir^2` (area of the base and the lid)

The problem provides the information that the volume of cylinder is of `4000 cm^3` .

You need to write the formula of volume of cylinder such that:

`V = pi*r^2*h =gt 4000 = pi*r^2*h`

You need to express the formula of TSA in terms of radius only, hence, using the formula of volume, you may write the height such that:

`h = 4000/(pi*r^2)`

You need to substitute `4000/(pi*r^2) ` for h in formula of TSA such that:

`TSA = 2*pi*r*4000/(pi*r^2) + 2pir^2 `

`TSA = 8000/r + 2pir^2`

You need to find the minimum surface area, hence you should differentiate the function TSA with respect to r such that:

`TSA' = -8000/(r^2) + 4pi*r`

You need to determine the dimensions of the cheapest container, hence you should introduce the information provided by problem with regard to cost such that:

`TSA' = (-8000/(r^2))*(0.0025)+ 4pi*r*0.005`

`TSA' =-20/(r^2) + 0.062*r`

You need to solve for r the equation TSA' = 0 such that:

`TSA' = 0 =gt -20/(r^2) + 0.062*r = 0`

`0.062*r^3 - 20 = 0 =gt r^3 = 20/0.062 =gtr = root(3)(322.580)=gtr = 6.858 cm`

`h = 4000/(pi*r^2) =gt h = 4000/(3.14*6.858^2)`

`h = 27.081 cm`

Hence, evaluating the dimensions of the cheapest container yields `r = 6.858 cm`  and `h = 27.081 cm.`

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