Fuel Tank Problem

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JSmith

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A fuel tank is being designed to contain 200 m3 of gasoline, but the maximum length of a tank (measured from the tips of each hemisphere) that can be safely transported to clients is 16 m long. The design of the tank calls for a cylindrical part in the middle, with hemispheres at each end. If the hemispheres are twice as expensive per unit area as the cylindrical part, find the radius and height of the cylindrical part so that the cost of manufacturing the tank will be minimal. Give your answers correct to the nearest centimetre.

How would I go about modelling and solving this problem? If someone could show me how they would go about solving it, that would be best. There are about 10 more similar problems so this would help me solve them. Thanks!!
 
There are different approaches that one could take. I would start by determining an expression for h (that's the length of the cylindrical section of the tank) in terms of the radius r.

The volume of the two hemispherical caps combine to make a sphere, yes? The volume of that sphere plus the volume of the cylindrical section must add up to the total volume given. Write that equation, and then solve it for h in terms of r.

Next, calculate the surface area of the sphere and of the cylinder because -- in this exercise -- minimizing cost means minimizing area. Double your result for the sphere's area (because the hemispheres' cost is double) and add that to the surface area of the cylindrical section. Use the first derivative with respect to r, in order to minimize the resulting expression for area.

Once you know the radius that minimizes area, you may substitute it into your expression for h, to find the length of the cylindrical part.

Check that your value for h satisfies the length constraint (i.e., 2r+h<=16).

Round your dimensions to two decimal places, so as to report to the nearest centimeter.
 
A fuel tank is being designed to contain 200 m3 of gasoline, but the maximum length of a tank (measured from the tips of each hemisphere) that can be safely transported to clients is 16 m long. The design of the tank calls for a cylindrical part in the middle, with hemispheres at each end. If the hemispheres are twice as expensive per unit area as the cylindrical part, find the radius and height of the cylindrical part so that the cost of manufacturing the tank will be minimal. Give your answers correct to the nearest centimetre.

Answer: r = 230cm
h is roughly 900cm

How do I get to these answers?
 
Surface Area

Cylindrical: \(\displaystyle 2\pi r h\)

2xHemisphere = 1 Whole Sphere: \(\displaystyle 4\pi r^{2}\)

Other

Length: \(\displaystyle 2r + h\)

Cost: \(\displaystyle 2\pi r h + 2\left(4\pi r^{2}\right)\)
 
tkhunny said:
Surface Area

Cylindrical: \(\displaystyle 2\pi r h\)

2xHemisphere = 1 Whole Sphere: \(\displaystyle 4\pi r^{2}\)

Other

Length: \(\displaystyle 2r + h\)

Cost: \(\displaystyle 2\pi r h + 2\left(4\pi r^{2}\right)\)
Click to expand...
I would add two more facts to this list of "what you know."

Length: \(\displaystyle 2r + h\ \ \ \le 16 m \)

Volume: \(\displaystyle \dfrac{4}{3} \pi r^3 + \pi r^2 h = 200 m^3 \) PLEASE CHECK THIS NUMBER!

What can you do with those formulas?
 
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DrPhil said:
I would add two more facts to this list of "what you know."

Length: \(\displaystyle 2r + h\ \ \ \le 16 m \)

Volume: \(\displaystyle \dfrac{4}{3} \pi r^3 + \pi r^2 h = 200 m^3 \) PLEASE CHECK THIS NUMBER!

What can you do with those formulas?
Click to expand...

I'm really unsure of how to start this, if someone could give me some steps, that would be awesome.
 
JSmith said:
I'm really unsure of how to start this, if someone could give me some steps, that would be awesome.
Click to expand...
It is not clear to me whether you need help translating the problem from words into math or in doing the pure math problem.

I am going to start with basics.

(1) Draw a sketch of what you are thinking about if you can. Many people find a sketch helps them think concretely about a problem.

(2) IN WRITING, assign a letter to each relevant variable. This takes a huge burden off your memory and helps organize the information. In differential calculus problems, I like to start with what is to be optimized.

c = cost (what is to be optimized).

u = surface area of cylinder.

v = surface area of hemisphere.

x = volume of cylinder.

y = volume of hemisphere.

z = volume of container.

h = height of cylinder

r = radius of cylinder and hemispheres.

(3) Write down what you know about the problem to be solved. Some of this information is general information that you are expected to know (such as formulas for the area and volume of spheres and cylinders) and some is given in the problem itself (such as the volume of the container is 200m^3.)

So z = 200. (Like Dr. Phil I wonder if this number is correct.)

And z = x + 2y.

(4) This method of attack generates a lot of variables so start eliminating variables. For this problem you will find that you can eliminate all the variables except c, h, and r. Actually, I suspect (but have not worked it out) that you can reduce it to two variables, c and h or c and r.

(5) You now have a pure math problem to solve, in this case a minimization problem subject to a constraint. What techniques do you know for solving such problems?

You may think this is unhelpful. If the issue is translating the problem into mathematical form, take it as far as you can so we can see where you are getting stuck. If the issue is that you do not know how to proceed once you have it translated, let's see your translation so we can give you a technique for solving.
 
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JeffM said:
It is not clear to me whether you need help translating the problem from words into math or in doing the pure math problem.

I am going to start with basics.

(1) Draw a sketch of what you are thinking about if you can. Many people find a sketch helps them think concretely about a problem.

(2) IN WRITING, assign a letter to each relevant variable. This takes a huge burden off your memory and helps organize the information. In differential calculus problems, I like to start with what is to be optimized.

c = cost (what is to be optimized).

u = surface area of cylinder.

v = surface area of hemisphere.

x = volume of cylinder.

y = volume of hemisphere.

z = volume of container.

h = height of cylinder

r = radius of cylinder and hemispheres.

(3) Write down what you know about the problem to be solved. Some of this information is general information that you are expected to know (such as formulas for the area and volume of spheres and cylinders) and some is given in the problem itself (such as the volume of the container is 200m^3.)

So z = 200. (Like Dr. Phil I wonder if this number is correct.)

And z = x + 2y.

(4) This method of attack generates a lot of variables so start eliminating variables. For this problem you will find that you can eliminate all the variables except c, h, and r. Actually, I suspect (but have not worked it out) that you can reduce it to two variables, c and h or c and r.

(5) You now have a pure math problem to solve, in this case a minimization problem subject to a constraint. What techniques do you know for solving such problems?

You may think this is unhelpful. If the issue is translating the problem into mathematical form, take it as far as you can so we can see where you are getting stuck. If the issue is that you do not know how to proceed once you have it translated, let's see your translation so we can give you a technique for solving.
Click to expand...

Just wanted first to confirm that the 200m^3 number is correct.

Now, I am thinking I need to put one of the length or volume formulas in terms of a variable, such as rearranging the length formula to: h
≤ 16m-2r.

Then i substitute this into the volume formula? I understand the actual formulas, I just need help with what steps to take to actually apply and solve this.
​
 
JSmith said:
Just wanted first to confirm that the 200m^3 number is correct.

Now, I am thinking I need to put one of the length or volume formulas in terms of a variable, such as rearranging the length formula to: h
≤ 16m-2r.

Then i substitute this into the volume formula? I understand the actual formulas, I just need help with what steps to take to actually apply and solve this.
​
Click to expand...
ok - I worked through the problem and V=200m^3 works fine. I was misreading your answers in cm, as if they were mm (that is one of the reasons I insist that units be included with EVERY physical quantity).

The 16-m constraint on the length is something you should check as a last step. If the answer you get optimizing cost doesn't work, only then would you use the length to determine h and r.

The first thing to try is using Volume to solve for h as a function of r. Use that relation to eliminate h in the Cost equation. I think we are all assuming that once you see that equation, you will know to differentiate and set the derivative to zero to find the radius for minimum cost.
 
Hi,

So they are asking to us optimize cost in terms of the surface area of the tank.

The equation for the surface of the tank is the surface of a sphere (2 hemispheres put together) and the surface area of the cylinder (ignoring the circle top and bottom);

SA=4Pir^2+2Pirh

Now to make the cost equation we just have to multiply the surface ara of the sphere by two, indicating the cost:

C = 8Pir^2+2Pirh

We can see we have 2 variables so we need to make a second equation. We are given the volume, 200m^3, so we can create the equation for the volume of the tank by adding the volume of the cylinder plus the volume of a sphere to give:

200 = 4/3Pir^3 + Pir^2h

Solve for h so we can input it in the cost equation, giving us only one variable;

h = (600 - 4Pir^3)/3Pir^2

Input this equal into the h in the cost equation, giving us;

C = 8Pir^2+2Pir((600 - 4Pir^3)/3Pir^2)

find the derivative of it and set it equal to zero. Your final derivative should be;

C' = 93Pir^3 - 3600

r should be 2.3 m, so 230 cm. Put 2.3 into the h equation to find h to be 8.96 m or 896 cm, which can be rounded to 900 cm.

The 16 m length of the tank is for the maximum and minimum, which we can ignore.

I hope this helps
 
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