**Question: Cost of producing cylindrical can determined by materials used for wall and end pieces. If end pieces are 3 times as expensive per cm**

^{2}as the wall, find dimensions (to nearest millimeter) to make a can at minimal cost with volume of 600 cm^{3}.**Relevant equations: a) V=600cm**

b) surface area = 2[pi]r

^{3}=[pi]r^{2}hb) surface area = 2[pi]r

^{2}+ [pi]rh**The miraculous attempt:**

a) solve h in terms of [pi]r

b) find surface area equation using h: SA=2[pi]r

c) first derivative: SA'= 4[pi]r - (600/r

a) solve h in terms of [pi]r

^{2}: 600=[pi]r^{2}h ; 600/[pi]r^{2}=hb) find surface area equation using h: SA=2[pi]r

^{2}+ 2[pi]r(600/[pi]r^{2}; SA=2[pi]r^{2}+ 2(600/r) ; SA=2[pi]r^{2}+ 1200/rc) first derivative: SA'= 4[pi]r - (600/r

^{2}) = 0I am regrettably hitting a brick wall on what should happen next. Additional information is that I

*think*the cost looks something like: 2[pi]r

^{2}= (2[pi]rh)(3), with the multiplication by 3 being the fact that the round end pieces are three times as expensive.

Any and all further help would be greatly appreciated.

Dan

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