2

You are in charge of manufacturing the snazzy new mobile tablets that everyone wants to own. The revenue function, in dollars, is given by

$R(s,t) = 8s+6t-s^2-2t^2+2st$ , s denotes "steel" model and t denotes "titanium" model, both in units of million (assume that you make positive but a finite number of products).

I have to determine the quantity of both products for maximum revenue.

My understanding:

So, I think the question is asking for the global maximum point. I found the critical point and it has only one, (11,7). Now, I think we need to assume that the lowest boundary for s and t is 0 and the upper boundary is also something (I don't know what to assume). And I'm stuck here.

kronos
  • 185

1 Answers1

2

I found the critical point and it has only one, (11,7).

This suggests trying to isolate relevant squares from the quadratic, and indeed a straightforward manipulation leads to the following form, which shows that there is one unique global maximum at $(11,7)$ without requiring any further assumptions: $\;R(s,t) = 65 - (t-7)^2 -(s - t - 4)^2\,$.

dxiv
  • 76,497
  • I still don't understand how you said that it is unique global maxima. Could you please explain it or tell me a site where I can read about it? – kronos Jun 29 '18 at 23:28
  • @kronos The square of a (real) number is non-negative i.e. $x^2 \ge 0$ with equality iff $x=0$. It follows that $R(s,t) = 65 - (t-7)^2 -(s - t - 4)^2 \le 65$ with equality iff $t-7=0$ and $s-t-4=0$. – dxiv Jun 29 '18 at 23:31