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dandaman
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Question: Cost of producing cylindrical can determined by materials used for wall and end pieces. If end pieces are 3 times as expensive per cm2 as the wall, find dimensions (to nearest millimeter) to make a can at minimal cost with volume of 600 cm3.
Relevant equations: a) V=600cm3=[pi]r2h
b) surface area = 2[pi]r2 + [pi]rh
The miraculous attempt:
a) solve h in terms of [pi]r2: 600=[pi]r2h ; 600/[pi]r2=h
b) find surface area equation using h: SA=2[pi]r2 + 2[pi]r(600/[pi]r2 ; SA=2[pi]r2 + 2(600/r) ; SA=2[pi]r2 + 1200/r
c) first derivative: SA'= 4[pi]r - (600/r2) = 0
I am regrettably hitting a brick wall on what should happen next. Additional information is that I think the cost looks something like: 2[pi]r2 = (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
Relevant equations: a) V=600cm3=[pi]r2h
b) surface area = 2[pi]r2 + [pi]rh
The miraculous attempt:
a) solve h in terms of [pi]r2: 600=[pi]r2h ; 600/[pi]r2=h
b) find surface area equation using h: SA=2[pi]r2 + 2[pi]r(600/[pi]r2 ; SA=2[pi]r2 + 2(600/r) ; SA=2[pi]r2 + 1200/r
c) first derivative: SA'= 4[pi]r - (600/r2) = 0
I am regrettably hitting a brick wall on what should happen next. Additional information is that I think the cost looks something like: 2[pi]r2 = (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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