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April 18, 2015

April 18, 2015

Posted by **Mary** on Sunday, November 30, 2008 at 12:09pm.

- calculus -
**Reiny**, Sunday, November 30, 2008 at 12:37pmlet the radius of the can be r mm

and its height be h mm

area = 1 rectangle + 2 circles

= 2pi(r)(h) + 2pi(r^2)

seems length = 2 circuferences + height of can

= 2(2pi(r)) + h

= 4pi(r) + h

so Cost = 2[2pi(r)(h) + 2pi(r^2)] + 1[4pi(r) + h]

but pi(r^2)h = 1000

so h = 1000/(pi(r^2))

this should give you a good start, sub h back into the Cost equation,

find the derivative of Cost,

then solve for r

sub back into Cost equation to find minimum cost

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