text: # Thread: More optimization...
  
  1. ## More optimization...
  
  Given C(x) = 680 + 4x + 0.01x^2 and demand is p(x) = 12 - x/500, find the value of x that will maximize profit.
  
  I assume that I somehow need to find the minimum for C(x) and the max for p(x), but I'm not sure if I somehow have to combine the two or something, or if the x has to be the same for both functions...
  
  2. If you think about it, profit is Revenue minus Cost. Also, Revenue must be Demand times Price. Mysteriously, you don't have a price, but your revenue is p instead of the more usual q. Hoping that p really is your demand and not a price you haven't included, I'm going to call it q, and call the price A. Then,
  
  $P\ =\ R - C\ =\ Aq - C$
  
  Just in case a picture helps...
  
  ... where straight continuous lines differentiate downwards (integrate up) with respect to x.
  
  Spoiler:
  
  Put dP/dx = 0 and then make x the subject in the bottom row.
  
  _________________________________________
  
  Don't integrate - balloontegrate!
  
  Balloon Calculus; standard integrals, derivatives and methods
  
  Balloon Calculus Drawing with LaTeX and Asymptote!
  
  3. So x = -50A/500 - 200?
  
  4. That was my conclusion, but could you post the whole problem? I think demand should depend on price, not quantity (i.e. demand is quantity, hence q), so something's missing after all.
  
  5. I did post the whole problem. That was all there was. "Given C(x) = 680 + 4x + 0.01x^2 and demand is p(x) = 12 - x/500, find the value of x that will maximize profit."
  
  With just the cost function and the demand function given I'm not sure how to turn those into the profit function and then get the answer.
  
  6. I suppose our result is ok as long as x isn't quantity sold. It could be, say, number of pages in a book, which might affect the cost and the demand.
  
  7. True. It gives no context so I assume that's what it meant...
  
  8. On the other hand, if the max profit is going to be for a value of x other than zero, then x must go less than zero, so pages wouldn't work!
  
  9. There has to be something we're missing...I mean, it has to have an actual value right?
  
  10. Okay, so I confirmed that the answer we came to isn't correct. Does anyone know what we did wrong?
  
  11. Okay, so a minimum value for the C(x) function is -200, which would be the value of x that minimizes cost the most, right? Well, I can't find it for p(x), on top of the fact that that doesn't answer the question of how to maximize the profit without a profit function....help?
  
  12. Bump...-200 isn't correct either.