1) You have made a common error. You have failed to define your terms. What is 'x'? What is 'y'?
2) You may be short some information, but if you live in an ideal world, feel free to assume that you won't need it and hope it will work out. If nothing else, this will encourage you to continue your exploaration, rather than giving up.
Watch these Definitions:
D = Diameter of Each Sphere. Spheres in any bucket are all the same size.
V = Volume of an individual Sphere
\(\displaystyle V\;=\;\frac{4}{3}\pi \left(\frac{D}{2}\right)^{3}\)
BV = Volume of the Bucket that we get to use to store spheres. Note, that this is not ALL the volume of the bucket. Spheres do not fit together perfectly. Also, we are about to assume that different sized spheres waste the same amount of space. This is a false assumption, but perhaps it will get us somewhere.
n = Number of Spheres in the Bucket = BV / V
Cost = Cost of One Sphere
\(\displaystyle Cost\;=\;k\cdot V^{2}\)
'k' is a constant of proportionality.
Total = Cost of an entire bucket of spheres.
Total = n*Cost
Note: Please notice that I did not spare the horses on this part of the problem. I have defined everything I think I need. I have discussed problems that I think might restrict our success. Most importantly, I just kept writing without worrying whether or not I actually would be able to solve the problem.
We are given, for D = 0.50, Total = 25.00. We'll need Total as a function of D. It will take some algebra.
\(\displaystyle Total(D)\;=\;n\cdot Cost\;=\;n\cdot k\cdot V^{2}\;=\;\frac{BV}{V}\cdot k\cdot V^{2}\;=\;k\cdot BV\cdot V\;=\;k\cdot BV\cdot \frac{4}{3}\pi \left(\frac{D}{2}\right)^{3}\)
Just to reiterate the final result:
\(\displaystyle Total(D)\;=\;k\cdot BV\cdot \frac{4}{3}\pi \left(\frac{D}{2}\right)^{3}\)
I have to admit I'm a little puzzled that we still have BV hanging around. However, I am not willing to go all the way to discouragement. If I look at the formula for Total(D), and maybe squint my eyes a little, I can imagine that k*BV is my constant of proportionality, rather than just k. More accurately, of course, 'k' is the constant for a single sphere. k*BV is the constant for the entire bucket. This gives:
\(\displaystyle Total(0.50)\;=\;k\cdot BV\cdot \frac{4}{3}\pi \left(\frac{0.50}{2}\right)^{3}\;=\;25.00\)
Solving for the constant:
\(\displaystyle k\cdot BV\;=\;\frac{1200}{\pi}\)
And
\(\displaystyle Total(D)\;=\;\frac{1200}{\pi}\cdot \frac{4}{3}\pi \left(\frac{D}{2}\right)^{3}\;=\;200\cdot D^{3}\)
Again, I'm a little surprised that it simplified so nicely. In any case, we're pretty much done. We now have a general forumla for the Total cost of a bucket of spheres of diameter D with the price of each sphere proprtional to the square of its volume. (Just for the record, they probably are not for individual sale!)
The only thing lacking is a discussion on limitations.
A) We probably have to stick with integer numbers of spheres. This causes us to question exactly what 'n' is. It cannot quite be as defined above.
B) There are limits on D, as well. Quite obviously, D > 0, but does it have an upper limit? It seems rational to believe that n = 1 is the minimum number of spheres in the bucket. This suggests that D is limited to the size of the bucket. Too bad we don't know the size of the bucket.
C) We have not addressed the problem of different sized spheres wasting different amounts of space.
D) Related to C is the consideration of the top of the bucket. Can the spheres stick out or will we shave off the top?
If there is a lesson I would want you to learn, it is that it is okay to explore and to wonder. Did we solve the problem? I think so. More importantly, we had FUN! Well, I did. Did you?
