A 3000 cu. ft. industrial tank is to be constructed by adjoining two hemispheres to the ends of a circular cyl

Question

A 3000 cu. ft. industrial tank is to be constructed by adjoining two hemispheres to the ends of a circular cylinder. If it costs twice as much per a square foot of surface are to construct the hemispherical ends as it does the cylinder middle, what should the dimensions be in order to minimize the cost of construction?

Answer 1

Let:

L = length of the cylindrical portion

r = radius of the tank

k = unit cost of per square foot of the cylindrical portion

V = volume of the tank = 3000 cubic ft

C = total cost of construction of the tank

(1). V = pi r^2 L + (4/3) pi r^3

r^2L + (4/3) r^3 - V/pi = 0

2rLdr + r^2dL + 4r^2dr = 0

dr(2rL + 4r^2) = - r^2dL

(2). dr/dL = -r^2/(2rL + 4r^2) = - r/(2L + 4r)

(3). C = 2pi rLxk + 4pir^2 x 2k

2rL + 8r^2 - C/(pi x k) = 0

2Ldr + 2rdL + 16rdr = 0

dr(L +8r) = - rdL

(4). dr/dl = - r/(L + 8r)

Equating (2) and (4)

-r/(2L + 4r) = -r/(L + 8r)

L + 8r = 2L + 4r

(5). L = 4r

Substituting the value of L in terms of r in Eq(1);

3000 = pi r^2(4r) + (4/3) pi r^3

3000 = 4pi r^3 + (4/3) pi r^3 = (16/3) pi r^3

r = (3000/((16/3) pi) ^(1/3) = 5.6363ft

L = 4r = 22.545ft

Answer 2

Obviously a long and slender tank. The typical propane tank is proportioned this way.

Answer 3

well, lets start with radius of "r" and see where we get

the surface area of a sphere is given as S=4*pi*r^2

the surface of a cylinder is given as S= pi*dia(d)*height(h)

the volume of the two sphere ends together is the same as the volume of one sphere which is 4/3*pi*r^3

the volume of the cylinder is pi*r^2*h

we know that the total volume has to equal 3000 ft^3 so,

3000=pi*r^2*h+4/3 * pi*r^3

if we solve for h, we will have the only height possible for each chosen radius that will give 3000 ft^3 tank

(3000-4/3 *pi*r^3)/pi*r^2=h

before we go there we can calculate what size of sphere will have a volume of 3000 ft^3

this will tell us the shortest possible tank (essentially no cylinder).

4/3*pi*r^3=3000

r=8.95 if I did that right

so, the biggest possible radius (giving a zero length cylinder between the sphere halves) is just under 9 feet

Now if I was doing this for work, I would skip the math, and get out a spreadsheet and just run the surface calculations for the vessels with radii between 0 and 9.  We can try to do the math next to make the mathmaticians and teachers happy.

with a radius of 1 foot, the height has to be 715 ft to get the 3000 ft^3 volume

that gives us a sphere surface of 13 ft^2 and a cylinder surface of 4494 ft^2

if you make a relative cost of surface by multiplying 2*13and adding that to 4494 you get about 4520

so, given shown as a table of form

r,h,cost

that would be

1,715,4520

continuing through the other radii we get

2,177,2325

3,77,1670

4,41,1427

5,24,1371

6,14,1429

okay, we see that its going up again (and it continues to go up from there as you would expect)

remember, the more cylinder the more total surface (the plain sphere is the smallest surface case) but the more cylinder, the LESS expensive sphere surface.  Thats why there is minimum cost point.

okay, we know the winning answer is a radius between 4 and 6 so we can try some smaller increments on our spread sheet and see where the turning point is

about 4.89

so, the low cost vessel has a

radius=4.89

diameter=9.78

straight wall height=25.1

now lets see if we can do it the hard way



(3000-4/3 *pi*r^3)/pi*r^2=h

the relative sphere cost is the sphere surface times 2 plus the cylinder surface, that is

2*4*pi*r^2+pi*2*r*h

if you plug in the equation above that is solved for "h" in place of h in this expression

then you have the relative cost function all in terms of "r"

if you take the derivative with respect to r and set that equal to zero, you can solve for the "r" that will give the minimum cost with the problem stipulation

if I have done my arithmetic right it should come out just like the trial and error method above.

good luck

Answer 4

total volume=3000 ft^3=(4/3)(pi)r^3+(pi)r^2h.....h=2865/{{... let S=total surface..so S=(4pi)r^2+(2pi)rh=(4pi)r^2+(2pi)r(2865/... S1=twice surface area of hemisphere=surface area of a sphere and S2=surface area of cylinder where r is the radius and h=cylinder height...we know from the cost constraint S=2S1+S2 and dS/dr={dS/d(S1)}{d(S1)/dr}=0...

we know {dS/d(S1)}=2 and d(S1)/dr=d/dr{(4pi)r^2=(8pi)r...we also know dS/dr=d/dr{(4pi)r^2+(2pi)rh=(8pi)r+(2pi)...

=2(8pi)r or (8pi)r=(2pi)h or 4r=h or h=4r..or r=h/4..substituting r into the first equation,  3000=(4/3)(pi)r^3+(pi)r^2(4r)=(4/3)(pi)r... or r^3=179.2 or 5.64..so h=4r=4(5.64)=22.6...check 3000=(4/3)pi(5.64^3)+pi(5.64)^2(4(5.64))... can also check the surface area of sphere=(4)(pi)(5.64)^2=400 and surface of cylinder=2pi(5.64)(22.6)=800...cost ratio=800/400=2..check.

Answer 5

i'd have told u to attempt ur own Differential HW , but luckily u got plenty of answers , anyway it's all about diff. the area (lateral area ) of the cylinder with the Sphere , which has the smallest area for same volume & largest volume for same area

Answer 6

I have not calculated the derivative of the opptimized cost function or iterated, but from the two first replies which arrived at 4.89 and 5.23 ft, I would likey choose 5 ft since I think it will be cheaper to get heads at standard (round) dimensions ... I guess it will  be much more expensive to have them rolled at odd sizes.  Just like when you size piping: say you calculated the optimum diameter is 4.78 inches, you will choose either 4", 6" and the same for tickness: usually sch 40 or sch 80, not something intermediate ... just my opinion.