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

Find the minimum cost of producing a can that contains 1000mm^3 if the cost of the metals is 2 cents per mm^2 and the cost to weld the seams is 1 cent per mm. Assume three welds;one around the top, one around the bottom, and one down the side.

- calculus -
**Reiny**, Sunday, November 30, 2008 at 12:37pm
let 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

## Answer this Question

## Related Questions

- math/economics in calculus - The average cost of manufacturing a quantity q of a...
- Word Problem Cost minimization - Don Garcia needs at least 24 units of vitamin A...
- math - The average cost of manufacturing a quantity q of a good, is defined to ...
- calculus - a rectangular box is to have a square base and a volume of 20 ft ...
- Math - A can of worms for fishing costs $2.60. There are 20 worms in a can. What...
- Calculus,math - Constuct a rectangular box with a square base that holds a given...
- Math - A manufacturer produces certain devices. The fixed cost of production is...
- Pre-Calculus - A local photocopying store advertises as follows. " We charge 14 ...
- Math - If the total cost function for producing x lamps is C(x) = 90 + 36x + 0....
- math - Formerly 1/6 of a pie cost 20 cents.Now the price of 1/8 of a pie is 30 ...

More Related Questions