Spherical Optimization and beyond

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Homework Help Overview

The discussion revolves around optimizing the dimensions of a tent designed in the shape of a spherical cap, with a focus on minimizing material costs while maintaining a specific volume of 150 cubic feet. The problem involves understanding the relationships between the surface area and volume of the tent structure.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the equations for volume and surface area of the dome and explore how to express the surface area in terms of a single variable to facilitate minimization. Questions arise about the differentiation process and the implications of having a constant volume.

Discussion Status

Some participants have provided guidance on how to approach the problem by suggesting the differentiation of the surface area equation. There is an acknowledgment of the need to focus on minimizing the surface area while keeping the volume constant, although there is no explicit consensus on the next steps.

Contextual Notes

The original poster expresses uncertainty about how to begin the optimization process and the implications of differentiating the equations. The constraints of maintaining a specific volume while minimizing costs are central to the discussion.

Muon12
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I have a semi-project due tomorrow that basically asks the following question: If you are designing a tent in the shape of a spherical cap (a sphere with the lower portion sliced away by a plane) and the material used for the roof costs 2.5 times more per square foot than the material used for the floor, what should the dimensions of the tent be to minimize the cost of materials used? Additional info: the volume of the tent will be 150 cu ft.
Goal: To derive a formula for maximum cost efficiency while retaining the original volume and shape of the tent.

The most difficult part of this problem is simply knowing where to start. I realize that when optimizing, I have to keep the minimums and maximum values of my equation in mind. In this case, I want to use as little ceiling material as possible for the tent while making sure that it remains a dome-shaped tent. Formulas used in this problem are the "volume of a dome" equation which I believe is V=((pi*h)/6)(3r^2+h^2) and the surface area formula of a dome: S=pi(h^2+r^2).
I wish I knew where to go from this beginning point, because short of taking the derivatives of these functions (with respect to what though?), I don't know to do.
 
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All right. You have two equations in three variables (note: I haven't checked your equations, I'll just assume they're right).

[tex]V = \frac{\pi h}{6}(3r^2 + h^2)[/tex]
and
[tex]S = \pi(h^2 + r^2)[/tex]

Notice that these are two equations in three variables--not four--because V is constant and known! V = 150 ft^3.

So why don't you use these two equations to solve for S in terms of a single variable (it doesn't matter which one), and then minimize that equation?

cookiemonster
 
there is still one more thing...

I've solved for r now and have a formula for the surface area of the dome in terms of h only. Now I need to minimize the "ceiling area" of this 150 cu.ft. tent. With an equation: S=pi(h^2+[(300/pi*h)-(h^2/3)]) I can now differenciate the surface area equation. So, do I simply differenciate the equation and look for the min. value, or is there more?
 
It's as simple as you think it is. Just differentiate and find the minimum.

As for your other post, you mentioned (perhaps a typo) that you can solve for either the minimum volume or the minimum surface area from that equation. Just a reminder: the volume is constant. It can't be minimized any more than it is.

cookiemonster
 
:smile: [?] [zz)] :frown: :wink: [b(]

dose are berry berry funkee

te gusta
 

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