Project Setup:You work for a company that manufactures 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.You will do this in stages, calculating the optimal dimensions for the part of the situation described at each stage.
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 the unused material from these squares is waste. The remainder of the drum is formed from a rectangular sheet of steel, assuming no waste. Ignoring all costs other than material cost, find the dimensions of the drum to minimizethe costof materials (steel).
2.The drum has seams around the perimeter of the top and bottom, as well as a vertical seam where the edges of the rectangular sheet are joined to form the lateral surface. In addition to the material cost of the steel, it costs $2 per meter to weld these seams. Find the dimensions of the drum that will minimize the cost of production (materials and welding).
3.The cost of shipping each drum from your manufacturing plant to the oil company depends on both the surface area of the drum (which determines how much each drum weighs) and the sum of the diameter and height (which affects how many drums can be transported on one vehicle). This cost is $1 per square meter of surface area plus $0.50 per meter of diameter plus height. Find the dimensions of the drum that minimize the total cost of production and transportation.
actually, if the radius of the tank is r, the sheets are the diameter wide, so each end is cut from a sheet (2r)x(2r) = 4r^2, making the total material used
8r^2+2πrh
not sure why you switched to cm. All the given dimensions are in meters, so I'll stay with that.
πr^2h = 1, so h = 1/(πr^2)
The cost of the drum (including waste) is
c = 3(8r^2 + 2πrh) = 6(4r^2 + 1/r)
dc/dr = 6(8r - 1/r^2) = 6(8r^3-1)/r^2
dc/dr = 0 when r = 1/2
so the drum has radius 1/2 and height 4/π = 1.2732
Now we have the welds. We have two circles of radius r and a line of height h. So, now the total cost is 3*area + 2*(2*circumference+height)
c = 3(8r^2 + 2/r) + 2(4πr+1/(πr^2)) = 24r^2 + 8πr + 6/r + 2/(πr^2)
Now dc/dr = 8π + 48r - 6/r^2 - 4/(πr^3)
= 2/(πr^3) (24πr^4 + 4π^2r^3 - 3πr - 2)
dc/dr=0 when r = 0.4389
so, h = 1/(π*.4389^2) = 1.6524
Finally, we have to add shipping cost. That makes the total cost
c = 3(8r^2 + 2/r) + 2(4πr+1/(πr^2)) + 1(8r^2 + 2πrh) + 1/2 (2r+h)
= 3(8r^2 + 2/r) + 2(4πr+1/(πr^2)) + 1(8r^2 + 2/r) + 1/2 (2r+1/(πr^2))
= 32r^2 + (8π+1)r + 8/r + 5/(2πr^2)
That means
dc/dr = 64r + (8π+1) - 8/r^2 - 5/(πr^3)
= (64πr^4 + (8π^2+π)r^3 - 8πr - 5)/(πr^3)
so,
dc/dr=0 when r=0.4558 and h=1/(π*0.4558^2) = 1.5322
whew
check my math
v = πr^2 h = 1
so, h = 1/(πr^2)
The surface area is
a = 2πr^2 + 2πrh = 2πr^2 + 2/r
for minimum area, find r where da/dr=0
The seam length is 2*2πr + h
now the total cost is
c = 3*(2πr^2 + 2/r) + 2(4πr+1/(πr^2))
for minimum cost, find where dc/dr = 0
I'll let you figure the total cost including shipping, but the idea is the same.
C'mon back if you get stuck, and show what you did.
The first question
I have A= 2r^2+2(pi)rh
V= (pi)r^2h -----> h=1/(pi)r^2h
A=2r^2+2(pi)r(1/(pi)r^2)-------> 2r^2+2/r
minimize 2r^2+(2/r)
A'=4r-2/r^2-----> LCD (4r^3-2)/(r^2)
critical points at A'=0
2(2r^3-1)=0 ------>
r=cuberoot(1/2)= (1/2)^1/3
h=1/((pi)(1/2)^2/3
I'm stuck after that, please help!!
I was told to use this formula for the area= r^2 + r^2 +2πrh
You are correct. I forgot that the ends were cut from square sheets, so the waste has to be counted. Good catch.
Anyway, it's late. I'll pick up on this tomorrow if no one else chimes in.
h= 1000/πr^2
1 L(V)= 1000 cm^3
A=4r^2+2πrh
C= 3[4r^2 + 2πr x 1000/πr^2]---------> 3[4r^2 + 2000/r]
= 12r^2 +6000/r
C'= 24r -6000/r^2
=24r-6000/r^2=0
24(r-250/r^2)=0
24(r^3-250)/r^2)
r^3-250= 0
r=3√(250)
h= 1000/π(250)^2/3
What about part 2 and 3 i'm totally stuck
Thanks in advance!