Maximum and minimum 283: minimize cost

In summary: V=\pi r^2h+\frac{4}{3}\pi r^3V=2\pi r^2h+\frac{8}{3}\pi r^3In summary, the cost of the body, C, is $8\pi r^3$ if the material of the lower hemisphere costs twice as much as the material of the sides, and $24\pi r^3$ if the material of the upper hemisphere costs three times as much.
  • #1
leprofece
241
0
283) you want to build a volume "V" shaped geometric body torque limited cylindrical half-spheres. If the material lower semisphere costs twice as much as the material of the sides, and the material of the upper hemisphere costs three times, calculate the dimensions of the body more economic.

answer H = 6R and R = cubic root of ( 3V/22pi)

Ok
V = 2pir2H+4/3pir3

The cost Function where i think it is my mistake is

C = 4pir2H +4pir3
 
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  • #2
Re: max and minimun 283

I have no idea what you mean by " 'V' shaped geometric body torque limited cylindrical half-spheres." Is this in fact just a right circular cylinder capped at both ends by hemispheres? If so, you are making the same mistake you made in the boiler problem concerning the volume of this body.

I would let the cost per unit square of the sides of the cylindrical portion be $k$ (in whatever units of currency you choose). Then your cost function would be:

Total cost = k(area of cylindrical portion) + 2k(area of lower hemisphere) + 3k(area of upper hemisphere)

You should now be able to state your objective function and constraint. What do you find?
 
  • #3
Re: max and minimun 283

MarkFL said:
I have no idea what you mean by " 'V' shaped geometric body torque limited cylindrical half-spheres." Is this in fact just a right circular cylinder capped at both ends by hemispheres? If so, you are making the same mistake you made in the boiler problem concerning the volume of this body.

I would let the cost per unit square of the sides of the cylindrical portion be $k$ (in whatever units of currency you choose). Then your cost function would be:

Total cost = k(area of cylindrical portion) + 2k(area of lower hemisphere) + 3k(area of upper hemisphere)

You should now be able to state your objective function and constraint. What do you find?

Cost = k(4pir2·+2k(2/3pir3)+3k 2k(2/3pir3)
This must be the cost equation
 
  • #4
Re: max and minimun 283

leprofece said:
Cost = k(4pir2·+2k(2/3pir3)+3k 2k(2/3pir3)
This must be the cost equation

No, what is the formula for the lateral surface of a cylinder? And why does your third term have factors of $2k$ and $3k$? Look at the formula I gave and carefully insert the correct area formulas.
 
  • #5
Re: max and minimun 283

MarkFL said:
No, what is the formula for the lateral surface of a cylinder? And why does your third term have factors of $2k$ and $3k$? Look at the formula I gave and carefully insert the correct area formulas.
C= 2pirh+ 2pir2 +2pir2
This?
 
  • #6
No. You have the correct lateral surface area of a cylinder, and the correct surface areas of the hemispheres, but you are not taking into account the differing costs per unit area. You want:

\(\displaystyle C(r,h)=k(2\pi rh)+2k\left(2\pi r^2 \right)+3k\left(2\pi r^2 \right)=2k\pi r\left(h+5r \right)\)

Now, can you state the constraint, and solve it for $h$ so that you will be able to express the objective function in terms of just the variable $r$?
 
  • #7
MarkFL said:
No. You have the correct lateral surface area of a cylinder, and the correct surface areas of the hemispheres, but you are not taking into account the differing costs per unit area. You want:

\(\displaystyle C(r,h)=k(2\pi rh)+2k\left(2\pi r^2 \right)+3k\left(2\pi r^2 \right)=2k\pi r\left(h+5r \right)\)

Now, can you state the constraint, and solve it for $h$ so that you will be able to express the objective function in terms of just the variable $r$?

Ok is it ?
V = 2pir2H+4/3pir3
 
  • #8
leprofece said:
Ok is it ?
V = 2pir2H+4/3pir3

If I am reading this correctly as:

\(\displaystyle V=2\pi r^2h+\frac{4}{3}\pi r^3\)

then no, that does not described the volume of the body. You have made this same error three times now. (Lipssealed)

You want:

\(\displaystyle V=\pi r^2h+\frac{4}{3}\pi r^3\)

Recall that the volume of a cylinder is $\pi r^2h$. Now, solve this constraint for $h$ and then substitute for $h$ into the cost function, and then minimize.
 

Related to Maximum and minimum 283: minimize cost

1. What is the significance of finding the maximum and minimum of 283 in terms of cost?

The maximum and minimum values of 283 in terms of cost can indicate the highest and lowest possible costs associated with a particular project or process. Finding these values can help in decision making and cost optimization.

2. How can the maximum and minimum of 283 be calculated?

The maximum and minimum values of 283 can be calculated by analyzing the cost function or data points related to the project. Various mathematical techniques such as differentiation, optimization, and regression can be used to find these values.

3. Can the maximum and minimum of 283 change over time?

Yes, the maximum and minimum of 283 can change over time depending on various factors such as changes in market conditions, inflation, and technological advancements. It is important to regularly review and update these values.

4. How can the maximum and minimum of 283 be used in cost management?

The maximum and minimum values of 283 can be used in cost management to set budget targets, identify areas of cost reduction, and track the progress of cost optimization efforts. They can also help in forecasting future costs and making strategic decisions.

5. Are there any limitations to using the maximum and minimum of 283 in cost analysis?

Yes, there can be limitations to using the maximum and minimum values of 283 in cost analysis. These values may not accurately reflect the actual costs in real-world scenarios due to various assumptions and simplifications made in the cost analysis process. It is important to consider these limitations while interpreting and using these values.

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