Optimization Problem: fuel tank for fighter jet that will hold 300 liters of jet fuel

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Hharris8

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Hello, newbie here. I need help bad y'all. I was given an optimization problem and it is due Tomorrow. I thought I had it figured out so I was about to turn it in until another classmate said the answer was different due to adding the area of the spheres on the end using 2/3piR^3. They did not know how to solve it, but knew the answer was around 21.07 or 20.07. I need a detailed explanation of how to get the answer. I have been working on it for hours and cannot figure it out and I have 5 finals this week. Thank you in advance!!

The question is "you are asked to design a fuel tank for a fighter jet that will hold 300 liters of jet fuel. The shape of the tank is a tube with rounded ends (this is where the spheres are) a cylinder with two hemispheres attached to each end. The side (the cylinder part) cost $.1 cm^2 to make while the rounded tops (hemisphere) cost $.3 per cm^2 to make. You will find out the radius of the tube that will minimize the manufacturing cost.

Find optimal radius and show your work
explain each step
explain how solution would change with the change in the total amount of fuel
 
Hharris8 said:
… I was given an optimization problem … I thought I had it figured out …

The question is "you are asked to design a fuel tank for a fighter jet that will hold 300 liters of jet fuel. The shape of the tank is a tube with rounded ends (this is where the spheres are) a cylinder with two hemispheres attached to each end. The side (the cylinder part) cost $.1 cm^2 to make while the rounded tops (hemisphere) cost $.3 per cm^2 to make. You will find out the radius of the tube that will minimize the manufacturing cost.

Find optimal radius and show your work
explain each step
explain how solution would change with the change in the total amount of fuel
Click to expand...
Okay. Please show us your steps and conclusions. We can start from there.

Please also read the forum guidelines. Thank you. :cool:
 
The steps I have so far, though I do not know if they are correct are:

h=(900-4pir^3)/3pir^2
v=4/3*pir^3 +pir^2h=300000
surface area = 4pir^2 + 2pirh
cost = 30(4pir^2) + 10(2pirh)

I then plug h into cost to get 30*(4pir^2) + 10*(2pir*((900-4pir^3)/3pir^2))
simplifying I get 120pir^2 + 20pir*((900-4pir)/3)

Then taking first derivative I get 240pir +20pi*(-4pi/3)

simplifying I get 240pir - (80pi/3)

further simplification 240pir= 80pi/3

further 240pir= 83.78

further r=.11

the answer for r is somewhere around 20-21 so I know this is wrong.
 
Looks to me like arithmetic and organization will help the most. I'm quite curious why you chose to report the expression for h BEFORE your formula for Volume. Backwards, it seems.

Please review how you went from your correct volume formula to your incorrect expression for h.
It should help if you fix that.

Note: There is NOT an operation known as "plug in". Try substitution, instead.
 
Last edited:
New h = 300/pir^2 - 4pi/3

for some reason v has =300000 where it should equal 300. Not sure how I typed it that way. Sorry about that.
 
Hharris8 said:
The steps I have so far, though I do not know if they are correct are:

h=(900-4pir^3)/3pir^2

v=4/3*pir^3 +pir^2h=300000

surface area = 4pir^2 + 2pirh

cost = 30(4pir^2) + 10(2pirh)

I then plug h into cost …

Then taking first derivative I get 240pir +20pi*(-4pi/3)

simplifying I get … r = .11
Click to expand...
Four comments:

1) I agree with TK; your expression for h is not correct.

2) The given costs are $0.30 and $0.10, not $30 and $10.

3) Most instructors will be looking for the derivative set equal to zero. Show this equation explicitly, before solving for r.

4) pir is not standard notation for pi*r

Your linear thinking looks good. :cool:
 
So, what I have gotten is 4/3*π*r^3 + π*r^2h

solving for h I get (300/π*r^2)- 4r/3

Substituting this into the equation along with the prices I get 1.2/3*π*r^3 + .1π*r^2 (300/π*r^2 - 4r/3)

Taking the derivative with repeat to r I get 1.2π*r^2 + .2π*r(300/πr^3 -4/3)

Setting this equal to zero I get r=1.36

Again, I know this is wrong.
 
I think I finally have gotten it figured out. The only question I have now is which formula would I plug r back into to prove it was a minimum and how would a change in volume change my answer? Do I just substitute V in place of 300 for the latter question? Thanks everyone!!
 
Whoa!! How did we get back down to 300? We need cc, not l. Your costs are in cm^2.
 
Hharris8 said:
So, what I have gotten is 4/3*π*r^3 + π*r^2h
Click to expand...

No, this is what you had gotten:

4/3*pi*r^3 + pi*r^2*h = 300000

At least, that's what you posted earlier.


solving for h I get 300/(π*r^2) - 4r/3
Click to expand...
This is almost correct.

h = 300000/(pi*r^2) - 4r/3

Cost function (decimal number rounded):

C(r) = 2.932153*r^2 + 60000/r
 
Hharris8 said:
I think I finally have gotten it figured out.
Click to expand...
What's your value for r?


… which formula would I plug r back into to prove it was a minimum …
Click to expand...
If you explain your steps, then setting C'(r)=0 and solving for r should be good enough.


… how would a change in volume change my answer?
Click to expand...
Try it and see.

That is, increase the volume from 300000 cm^3 to 400000 cm^3, and see what happens to r.

Try decreasing the volume, too; see what happens to r.
 
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