# Homework Help: Spherical Optimization and beyond

1. Mar 4, 2004

### Muon12

I have a semi-project due tommorrow 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.

2. Mar 4, 2004

### cookiemonster

All right. You have two equations in three variables (note: I haven't checked your equations, I'll just assume they're right).

$$V = \frac{\pi h}{6}(3r^2 + h^2)$$
and
$$S = \pi(h^2 + r^2)$$

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

3. Mar 5, 2004

### Muon12

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?

4. Mar 5, 2004

### cookiemonster

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

5. Mar 5, 2004

### radmonovic

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