Wikipedia:Reference desk/Archives/Mathematics/2010 November 18

= November 18 =

A financial math problem
Dear Wikipedians:

I've worked out part (a) of the following question, but feels that part (b) is more difficult and I'm not sure how to proceed

A manufacturer finds that when 8 units are produced, the average cost per unit is $64, and the marginal cost is $18. (a) Calculate the marginal average cost to produce 8 units (b) Find the cost function, assuming it is a quadratic function and the fixed cost is $400.

So my solution for (a) is

$$ \overline{C} = \frac{C}{q} \Rightarrow \frac{d\overline{C}}{dq} = \frac{\frac{dC}{dq}q - C\frac{dq}{dq}}{q^2}$$

when q = 8,

$$ \frac{d\overline{C}}{dq} = \frac{18 \times 8 - 8 \times 8^2}{8^2} = -5.75 $$

for part (b), I know that the final form of the function looks something like

$$ C(q) = a_2q^2 + a_1q + a_0 $$

where a0 is equal to 400. But I don't know how to get the other coefficients from the given information. So I need your help.

Thanks,

70.29.24.19 (talk) 01:33, 18 November 2010 (UTC)


 * You have $$C(8)$$ since you have $$\overline{C}(8)$$. You also have $$C'(8) = 18$$. 67.158.43.41 (talk) 04:22, 18 November 2010 (UTC)


 * Thanks! Got it! 70.29.24.19 (talk) 04:38, 18 November 2010 (UTC)

Simple extremal subgraph query
Hi, is there an explicit function in (n,s) for the maximal number of edges a graph on n vertices may have such that the degree of every vertex is less than s? WLOG assuming n$$>$$s - I thought perhaps it was to do with the floor function, certainly you can obtain $$(\lfloor\frac{n}{s}\rfloor \cdot \frac{s(s-1)}{2}) + \frac{(n-(\lfloor\frac{n}{s}\rfloor)s)(n-(\lfloor\frac{n}{s}\rfloor)s-1)}{2}$$, by just splitting the vertices up into classes of size at most s. However, is this necessarily an upper bound? What is the form of such a function? Estrenostre (talk) 04:39, 18 November 2010 (UTC)
 * The upper bound is if every vertex has degree exactly s-1, in which case you've got n(s-1)/2 edges. Obviously if n and s-1 are both odd, that value is not possible since it's not integer, so it needs to be $$\lfloor\frac{n(s-1)}{2}\rfloor$$.  I'm pretty sure this value is obtainable. Rckrone (talk) 11:01, 18 November 2010 (UTC)
 * It is indeed. Let d = s − 1. If d is even, define a graph on {0,...,n − 1} by connecting each vertex a by an edge with (a + i) mod n, where i = −d/2,...,−1,1,...,d/2. If d is odd, use the same vertex set, connect a with (a + i) mod n for i = −(d + 1)/2,...,−2,2,...,(d + 1)/2, and moreover, connect each odd a with a − 1 (and therefore each even a < n − 1 with a + 1).—Emil J. 13:09, 18 November 2010 (UTC)