*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
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
= 0
8π^2 r^3 = 4
r^3 = 1/(2π^2)
r = .37
h = 1/(π(.37^2)) = 2.325
check:
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
2.
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
2(2*2πr+h)
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.
3.
shipping costs
1(2πr^2 + 2πrh) + 1/2 (2r+h)
add that to production.
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?
looks like Steve had the same reservation about cost of material.
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
To solve this optimization problem, we need to find the dimensions of the drum that minimize the cost of production and transportation. Let's break down each part of the problem step by step:
1. Determining the dimensions of the drum that minimize the cost of production:
- We are given that the drum needs to have a volume of 1 cubic meter.
- Let's assume the radius of the drum is "r" and the height is "h".
- The volume of a cylindrical drum is given by V = π * r^2 * h.
- We need to minimize the cost, which includes material cost and welding cost.
- The material required can be calculated as the surface area of the drum, which is S = 2πr^2 + 2πrh.
- The cost of material is proportional to the surface area, so we can say cost of material = k * S, where k is some constant.
- The welding cost is $2 per meter of seam, and the length of seams is the perimeter of the top and bottom circles plus the length of the vertical seam, which is 2πr + 2h.
- The welding cost is therefore 2 * (2πr + 2h) = 4πr + 4h.
- The total cost can be represented as C = k * S + 4πr + 4h.
- We need to minimize C subject to the constraint V = 1 cubic meter.
- To do this, we can express C in terms of only one variable, either r or h, using the constraint equation.
- Substitute the expression for V into the equation C = k * S + 4πr + 4h, and solve for the remaining variable.
- Finally, find the critical points of the resulting equation, and determine which one minimizes the cost of production.
2. Determining the dimensions of the drum that minimize the cost of transportation:
- We are given that the cost of transportation depends on the surface area of the drum and the sum of diameter and height.
- Let's assume the radius of the drum is "r" and the height is "h".
- The surface area of the drum is given by S = 2πr^2 + 2πrh, as we calculated before.
- The cost of transportation can be represented as T = 1 * S + 0.50 * (2r + h), where 1 is the cost per square meter of surface area and 0.50 is the cost per meter of diameter plus height.
- We need to minimize T subject to the constraint V = 1 cubic meter.
- Similarly to the previous step, express T in terms of only one variable using the constraint equation.
- Substitute the expression for V into the equation T = 1 * S + 0.50 * (2r + h), and solve for the remaining variable.
- Find the critical points of the resulting equation, and determine which one minimizes the total cost of production and transportation.
These steps should help you solve the optimization problem and find the dimensions of the drum that minimize the cost of production and transportation.