Optimization

[DerivativesAreFun]

Hi,

I have a question concerning optimization and if my logic is correct on this problem or not.

Here is the problem:

Create a storage tank that with two hemispheres on a circular cylinder at a minimal cost.
Volume must = 20 pi
End pieces (hemispheres) cost 3 times the cylinder piece

So,

Volume = 4/3?r^3 + 2?rh = 20?
h = (20? - 4/3?r^3)/2?r
h = 10/r - 2/3r^2

Cost = 4?r^2(3x) + 2?rh(x) = ? .. where 3x represents the sphere, 3 times the cost of the cylinder, which is represented by x..
12x?r^2 + 2x?r (10/r - 2/3r^2)
12x?r^2 + 20x? - 4/3x?r^3
x?(12r^2 + 20 - 4/3r^3) ..factored out x?..

Derivative = ?(12r^2 + 20 - 4/3r^3) + x?(24r - 4r^2) ..product rule..
12?r^2 + 20? - 4/3?r^3 + 24x?r - 4x?r^2
?(12r^2 + 20 - 4/3r + 24xr - 4xr^2) ..factored out ?..

I'm not sure where I am going from here, or if I am even going in the right place. Could someone please help me out with this problem?

Thanks.

 

[tkhunny]

You pobably should start by fixing the Volume formula. You sem to have the lateral surface area of the cyllindrical portion in there. A little dimensional investigation would help avoid this. The degree of each VOLUME term should be 3.

I get

\(\displaystyle r = \frac{1}{7}\cdot\sqrt[3]{735}\)

 

[mmm4444bot]

Volume = 4/3?r^3 + 2?rh ? 2?rh is not the volume of a cylinder

Cost = 4?r^2(3x) + 2?rh(x)

We don't need to introduce symbol x; they don't ask for any actual cost.

Just multiply the end caps' surface area by 3.

C(r) = 12 ? r^2 + 2 ? r h

After all, it could be interpreted that the same material is used for both the hemispherical end caps and the cylinder body but that the endcaps must be three layers thick. Therefore, the end caps cost three times as much as the cylinder.

Otherwise, your reasoning seems good.

Find the correct expression for h in terms of r.

Determine C`(r).

Solve C`(r) = 0.

Thank you for showing your work.

My minimizing value for the radius matches tkhunny's.

Cheers ~ Mark

MY EDITS: Fixed typographical error that said to multiply the cylinder surface by 3; added confirmation of tk's answer

 

[soroban]

Hello, DerivativesAreFun!

Your geometry is a little off . . .

A storage tank has two hemispherical caps on a circular cylinder.
Its volume must be \(\displaystyle 20\pi\)
The end pieces (hemispheres) cost 3 times as much per square unit as the cylindrical piece.

Construct the storage tank with minimum cost.

\(\displaystyle \text{The volume is: }\;\tfrac{4}{3}\pi r^3 + \pi r^2h \;=\;20\pi \quad\Rightarrow\quad h \:=\:\frac{60-4r^3}{3r^2}\) .[1]

\(\displaystyle \text{The area of the two hemispheres is: }\,4\pi r^2\)
. . \(\displaystyle \text{At }3a\text{ dollars per square unit, its cost is: }\,12a\pi r^2\)

\(\displaystyle \text{The area of the cylinderical side is: }\,2\pi r h\)
. . \(\displaystyle \text{At }a\text{ dollars per square unit, its cost is: }\,2a\pi rh\)

\(\displaystyle \text{Hence, the total cost is: }\:C \;=\;12a\pi r^2 + 2a\pi rh \;=\;2a\pi(6r^2 + rh)\) .[2]


Substitute [1] into [2]:

. . \(\displaystyle C \;=\;2a\pi\bigg[6r^2 + r\left(\frac{60-4r^3}{3r^2}\right)\bigg] \;=\;2a\pi\bigg[6r^2 + \frac{60-4r^3}{3r}\bigg]\)

. . \(\displaystyle C \;=\;2a\pi\bigg[\frac{18r^3 + 60 - 4r^3}{3r}\bigg] \;=\;2a\pi\bigg[\frac{14r^3+60}{3r}\bigg] \;=\;\tfrac{4}{3}a\pi\bigg[\frac{7r^3 + 30}{r}\bigg]\)

. . \(\displaystyle C \;=\;\tfrac{4}{3}a\pi\left[7r^2 + 30r^{-1}\right]\)


And that is the function we must minimize . . .

 

[DerivativesAreFun]

Thanks for the quick responses, I appreciate it.

Let me make sure I got this right:

V = 4/3?r^3 + ?r^2h = 20?
h = (60 - 4r^3) / (3r^2)

SA = 4?r^2 + 2?rh

Cost = C(x) = 3(4?r^2) + 2?rh
C(x) = 12?r^2 + 2?r((60 - 4r^3) / (3r^2))
C(x) = 12?r^2 + (120?r - 8?r^4) / (3r^2))
C(x) = (36?r^4 + 120?r - 8?r^4) / (3r^2)
C(x) = (24?r^4 - 120?r) / (3r^2)

C'(x) = ((96?r^3 - 120?)(3r^2) - (24?r^4 - 120?r)(6r)) / ((3r^2)^2)
C'(x) = (288?r^5 - 360?r^2 - 144?r^5 + 720?r^2) / ((3r^2)^2)
C'(x) = (144?r^5 + 360?r^2) / ((3r^2)^2)
C'(x) = (48?r^3 + 120?) / (3r^2) ... (We can dispose of the 3r^2 since that number would be undefined and imaginary in the real world)

r^3 = (120?) / (48?)
r^3 = 2.5
r = 1.35720881

h = (60 - 4r^3) / (3r^2)
h = 9.04805869

C(x) = 3(4?r^2) + 2?rh
C(x) = 12?(1.35720881^2) + 2?(1.35720881)(9.04805869) = 146.600533

3x + x = 146.600533
4x = 146.600533
x = 36.6501333

Cost of cylinder = 36.6501333
Cost of hemispheres = 109.9504

Can both costs of each piece of the tank be varied by the use of x? so that:

Cost of cylinder = 36.6501333x
Cost of hemispheres = 109.9504x

Thanks.

 

[galactus]

The radius should be \(\displaystyle r=\frac{15^{\frac{1}{3}}\cdot 7^{\frac{2}{3}}}{7}\approx 1.28923\)

 

[DerivativesAreFun]

Okay, thanks for the response.

Let me try this one more time!!

Okay, so these are my results:

r = 1.289231989
h = 11.05055991

So...If I wanted to figure out cost, is this right?:
Cost of hemispheres = 12?r^2 = 12?(1.289231989^2) = 62.66041466
Cost of cylinder = 2?rh = 2?(1.289231989)(11.05055991) = 89.51487812

So, if I want to make this where price, x, can vary, do I write it like this?:
Cost of hemispheres = 62.66041466x
Cost of cylinder = 89.51487812x

How can this be right if the hemispheres cost 3 times the amount of the cost of the cylinder?

Thanks, sorry for not getting this the first time! :arrow: :? :?:

 

[mmm4444bot]

r = 1.289231989

I'm not sure why you're reporting these values to the 100-millionth place.

I don't care, yet I note that your rounding is not correct, on some of them.

I choose to round to four places versus eight.

r = 1.2892


h = 11.05055991 This is not correct.

I get h = 10.3139


So...If I wanted to figure out cost, is this right?:

Cost of hemispheres = 12?r^2 = 12?(1.289231989^2) = 62.66041466

This is not the cost; it's three times the surface area of the sphere (in square units).

Cost of cylinder = 2?rh = 2?(1.289231989)(11.05055991) = 89.51487812

That value is incorrect because h is incorrect. Try 83.5472, instead.

This is not the cost; it's the surface area of the cylinder (in square units).


So, if I want to make this where price, x, can vary, do I write it like this?:

Cost of hemispheres = 62.66041466x
Cost of cylinder = 89.51487812x

Yes, but fix those values, and explicity define the meaning of your symbol x.

x = the dollar cost (per square unit) of the cylinder material

C(x) = the cost of the tank

C(x) = 62.6604x + 83.5472x

This seems to jive with the cost formula posted by Soroban:

c(x) = 4/3 * x * Pi * [7r^2 + 30r^(-1)]

Using the values r = 1.2892 and x = 29.75 (dollars), I get:

C(29.75) = 4349.68

c(29.75) = 4349.68

 

[DerivativesAreFun]

Thanks mmm4444bot. Sorry about the place values, I'll try to be more careful next time. I also recalculated h and seemed to get your answer, not sure what I did the first time.

One last, final question:

Cost of cylinder = 83.5472x
Cost of hemispheres = 62.6604x

The TOTAL cost of the hemispheres does not have to be greater than the total cost of the cylinder, correct? But, the hemispheres still cost 3 times more per square foot than the cylinder, correct?

Thanks for all of your help!

 

[mmm4444bot]

Sorry about the place values

There is no need to be sorrowful. I noted the improper rounding to 8 places; I don't care how many places you want to report, only that values are properly rounded.

I mean, I don't know why you need that much precision, but it's okay. Just watch your rounding.

The TOTAL cost of the hemispheres does not have to be greater than the total cost of the cylinder, correct?

That's my interpretation. The sphere always costs less than the cylinder, for any x. This is because the cylinder requires about 90 square units of material and the sphere requires about 21 square units of material.

In other words, you need almost four and a quarter times more material to build the cylinder. So its total cost is higher.


But, the hemispheres still cost 3 times more per square foot than the cylinder, correct?

That's the way I read the exercise.

Going back to my example of $29.75 per square unit for cylinder material, we have the following.

4 Pi r^2 is the surface area of the sphere. With our optimized value of r, it works out to 20.8868 square units.

62.6604(29.75) = 1864.1469

The total cost of the sphere is $1,864.15 (rounded). Dividing that amount by the area gives $89.25 per square unit for the sphere material.

3 times $29.75 equals $89.25.

In other words, the unit-cost of the sphere's material is three times the unit-cost of the cylinder's material.

You wrote, "End pieces (hemispheres) cost 3 times the cylinder piece".

I read that to mean, "The end-piece material is three times the cost of the cylinder material".

If they want the total cost to build the sphere to equal three times the total cost to build the cylinder, then I misinterpreted their intent.

 

[DerivativesAreFun]

I interpreted the question the same way you did, the material and not the total cost.

Great explanations, thanks to everyone for all the help - especially mmm4444bot!

Thanks. :!: :!:

 

[mmm4444bot]

I scribbled, while eating YUM PLA MUCK and PAD PAK PASOM at the local Thai place.

If we want the total cost of the sphere to be three times the total cost of the cylinder, I think the minimizing value of r changes.

C(x) = 4 * Pi * r^2 * A + 2 * Pi * r * h * B

A = unit cost of sphere material

B = unit cost of cylinder material

I get r = 2.1086 and A = -2B + 30B/r^3

EG:

The cost of cylinder material is $29.75 per square unit.

That makes the cost of sphere material $35.70 per square unit.

The sphere cost is 4 * Pi * r^2 * A = 1994.6490

The cylinder cost is 2 * Pi * r * h * B = 664.8909

Three times $664.89 is $1,994.67 and that's close enough for me.

 

[DerivativesAreFun]

Alright, that's cool, now I know how to do it both ways!

Thanks.