*Optimization problem* I'm okay at some optimization problems, but this one has me stumped.
You work for a company that manufactures circular cylindrical steel drums that can be used to transport various petroleum products. Your assignment is to determine the dimensions (radius and height) of a drum that is to have a volume of 1 cubic meter while minimizing the cost of the drum.
2. The drum has seams around the perimeter of its top and bottom, as well as a vertical seam where edges of the rectangular sheet are joined to form the lateral surface. In addition to the material cost, the cost of welding the seams is $2 per meter. Find the dimensions of the drum that will minimize the cost of production.
3. The cost of shipping each drum from your plant in Birmingham to an oil company in New Orleans depends upon both the surface area of the drum (which determines weight) and the sum of the diameter and the height (which affects how many drums can be transported on one vehicle). The cost is estimated to be $1 per square meter of surface area plus $0.50 per meter of diameter plus height. Find the dimensions of the drum that will minimize the total cost of production and transportation.
I know, its a very lengthy problem, but any help is appreciated
Calculus - Reiny, Thursday, November 29, 2012 at 11:01am
Look carefully what we are minimizing
in #2, it says, "... minimize the cost of production"
so that will have to be our main equation
Let the radius of the drum be r, let its height be h
we are told that πr^2h = 1
h = 1/(πr^2)
length of seam = 2 circuferences + 1 height
= 2(2πr) + h
= 4πr + 1/(πr^2)
cost of production
= 2(4πr + 1/(πr^2) )
= 8πr + (2/π)r^-2
d(cost...)/dr = 8π - (4/π)r^-3
8π^2 r^3 = 4
r^3 = 1/(2π^2)
r = .37
h = 1/(π(.37^2)) = 2.325
1. for correct volume:
π(.37^2)(2.325) = .999945.. , not bad
2. take a value of r slight off on either side of .37
r = .35, then Cost = 13.993
r = .37 , then cost = 13.949 lowest of the three
r = .39 , then cost = 13.987
all looks good.
3. I will let you try that one, follow the kind of thinking I used in #2, keep the same definitions of r and h
Calculus - Steve, Thursday, November 29, 2012 at 11:02am
To minimize the cost of production, it is necessary to know the relative cost of the material and the seams.
If the material costs $m/m^2, then the cost of materials is
m(2πr^2 + 2πrh)
The cost of welding is
so, the total cost is
c(x) = 2πmr^2 + 2πmrh + 8πr + 2h
Now, πr^2h = 1, so h = 1/(πr^2), and
c(x) = 2πmr^2 + 8πr + 2m/r + 2/πr^2
dc/dx = 4mr + 8π - 2m/r^2 - 4/πr^3
As you can see, the cost of material has an effect on the optimal dimensions.
1(2πr^2 + 2πrh) + 1/2 (2r+h)
add that to production.
Calculus - Reiny, Thursday, November 29, 2012 at 11:04am
Rethinking my solution to #2, I feel that we should have included the material cost, that is, the surface area, in our Cost equation.
But you did not state a cost of the sheet metal, so .... ??
How does my answer match up with your textbook answer?
Calculus - Reiny, Thursday, November 29, 2012 at 11:07am
looks like Steve had the same reservation about cost of material.
Calculus - Zared, Thursday, November 29, 2012 at 3:11pm
My bad on that. I missed the first part. I thought I pasted it in. Here it is.
1. The cost of the steel used in making the drum is $3 per square meter. The top and bottom of the drum are cut from squares, and all unused material from these squares is considered waste (i.e. counted in the cost as well). The remainder of the drum is formed by a rectangular sheet of steel, assuming no waste. Ignoring all costs other than material cost, find the dimensions of the drum that will minimize the cost.
Thank you for all the help