An aluminum can is to be constructed to contain 2400 cm3 of liquid. Let r and h be the radius of the base and the height of the can respectively. a) Express h in terms of r. (If needed you can enter π as pi.) h= b) Express the surface area of the can in terms of r. Surface area = c) Approximate the value of r that will minimize the amount of required material (i.e. the value of r that will minimize the surface area). What is the corresponding value of h? r= h=
Volume V = pi r^2 h and Volume = 2400 cm^3 2400cm^3=pi r^2 h h = (2400)/(pi r^2) =764/r^2 That is part a. b. is same process except use formula for Surface Area
Surface Area SA= 2 pi r h + 2 pi r^2 includes the circular bases
oh thanks
Substitute the value you obtained in part a. for h in the surface area. Then use derivative to solve for r.
S = 2 pi r (764/r^2) + 2 pi r^2 now do dS/dr make that equal zero then solve for r.
Look at dumbcow solution for a similar problem. http://openstudy.com/study#/updates/4e376c520b8bf47d065ed7f6
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