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A cylindrical drum is made to hold exactly 1m^3 in its interior. Assume that the material for the top and the bottom costs $20...Asked by Frederique
                A cylindrical drum is made to hold exactly 1m^3 in its interior. Assume that the material for the top and the bottom costs $20 per m^2, while that for the side costs $10 per m^2. Determine the radius of the drum that minimizes the cost of the material used. 
____
/ \
|\____/| <--------Drum
| | h
|______|
I posted this question before but I wasn't sure about one part.
Eqn's:
(1) Area = 2PI*r^2 + 2PI*rh
(2) Cost = 20*2PI*r^2 + 2PI*rh
Solving for "h" for eqn (1)
h = -Area(2PI*r^2/(2PI*r)
did I do this step right cuz after i solve for h I am able to sub that into eqn 2, and find DCost/dt but I feel Im missing a step. Any help will be appreciated! :)
Use the volume equation to solve for h.
Volume= PI r^2 h
h= volume/PIr^2
Then put that into cost.
Cost = 20*2PI*r^2 + 2PI*r(volume/PIr^2)
Now simplify, and take the derivative of cost with respect to r.
            
            
        ____
/ \
|\____/| <--------Drum
| | h
|______|
I posted this question before but I wasn't sure about one part.
Eqn's:
(1) Area = 2PI*r^2 + 2PI*rh
(2) Cost = 20*2PI*r^2 + 2PI*rh
Solving for "h" for eqn (1)
h = -Area(2PI*r^2/(2PI*r)
did I do this step right cuz after i solve for h I am able to sub that into eqn 2, and find DCost/dt but I feel Im missing a step. Any help will be appreciated! :)
Use the volume equation to solve for h.
Volume= PI r^2 h
h= volume/PIr^2
Then put that into cost.
Cost = 20*2PI*r^2 + 2PI*r(volume/PIr^2)
Now simplify, and take the derivative of cost with respect to r.
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