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

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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

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

when q = 8,

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

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

$C(q)=a_2{q}^2+a_{1}q+a_0$

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 $\bar{C}(8)$. You also have $C^{\prime }(8)=18$. 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)