Optimization problem

[james23]

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 cost of 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.


This is what I have for the first question
A= 2r^2+2πrh
V= πr^2h -----> h=1/πr^2h
A=2r^2+2πr(1/(πr^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/π(1/2)^2/3

I'm stuck after that, please help!!

 

[Steven G]

Your 1st line is wrong. The area of the top and bottom are each d^2 (where d= diameter), not r^2.
Can you please write equal signs (=) instead of arrows!
Critical points of A are not only when A'=0. It is when A'=0, end points or undefined.
Now you have h_min. So use that to find r_min. Then calculate A_min, So what is the min Cost, C_min

 

[Dr.Peterson]

This is what I have for the first question
A= 2r^2+2πrh
V= πr^2h -----> h=1/(πr^2h)
A=2r^2+2πr(1/(πr^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/π(1/2)^2/3

I'm stuck after that, please help!!

Error 1: Each side of the squares is 2r, not r. (You are right to use the squares, not circles.)
Error 2: You didn't mean to have h on both sides, did you? But this disappears later.
Question 1: What do you mean by LCD? Ah, I see -- you meant that you used the LCD to combine the fractions.

Fix the two errors, and you're doing well; I like (most of) your arrows, as too many people use "=" when they mean arrows! And it's okay so far not to bring in the cost explicitly, as that is proportional to surface area used anyway. (But you should verify that what you get is really the minimum; as Jomo mentioned, the minimum could also occur at an end point of the domain.)

Now, on to part 2. Here you will have to explicitly write a cost function, using both area and seam length. Give it a try, and show us what you get.

 

[james23]

Error 1: Each side of the squares is 2r, not r. (You are right to use the squares, not circles.)
Error 2: You didn't mean to have h on both sides, did you? But this disappears later.
Question 1: What do you mean by LCD? Ah, I see -- you meant that you used the LCD to combine the fractions.

Fix the two errors, and you're doing well; I like (most of) your arrows, as too many people use "=" when they mean arrows! And it's okay so far not to bring in the cost explicitly, as that is proportional to surface area used anyway. (But you should verify that what you get is really the minimum; as Jomo mentioned, the minimum could also occur at an end point of the domain.)

Now, on to part 2. Here you will have to explicitly write a cost function, using both area and seam length. Give it a try, and show us what you get.

How about now?
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

2.
C = 12r^2 +6000/r + 4πr

3.

C = 12r^2 + 6000/r + 4πr + 2πrh+ 2πr^2 + r + .5h


Thank You for your help.

 

[Dr.Peterson]

How about now?
h= 1000/πr^2 <--- Why 1000? Everything here is in meters, and volume is 1 m^3.
1 L(V)= 1000 cm^3 <--- Where did it mention liters?
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) <--- This means 3 times the square root; use some notation for the cube root instead
h= 1000/π(250)^2/3

2.
C = 12r^2 +6000/r + 4πr <--- You omitted the vertical seam

3.

C = 12r^2 + 6000/r + 4πr + 2πrh+ 2πr^2 + r + .5h

See comments in red above.

 

[james23]

See comments in red above.

1.
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

2.
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

3.
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

 

[Dr.Peterson]

I haven't checked every detail, but I see you caught an error or two that I'd forgotten to point out, and have expressed your work very clearly. Did you use a numerical method (e.g. graphing calculator) to solve the cubic and quartic equations?

As far as I can tell, it looks good. If I wanted to be more sure of the actual answer, I might calculate the cost in each case that you found to be optimal, and then find costs for a couple other values of r near those, and confirm that the latter are larger.

 

[james23]

I haven't checked every detail, but I see you caught an error or two that I'd forgotten to point out, and have expressed your work very clearly. Did you use a numerical method (e.g. graphing calculator) to solve the cubic and quartic equations?

As far as I can tell, it looks good. If I wanted to be more sure of the actual answer, I might calculate the cost in each case that you found to be optimal, and then find costs for a couple other values of r near those, and confirm that the latter are larger.

Yes, I used a math software to solve the equations, and they checked out. Thanks for your help!