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
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.
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
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