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.
explanation: The document is a discussion thread about a math problem involving optimization. The problem is to find the value of \( x \) that maximizes profit given the cost function \( C(x) = 680 + 4x + 0.01x^2 \) and the demand function \( p(x) = 12 - \frac{x}{500} \). The thread includes a question, multiple attempts at solving the problem, and a discussion about the correctness of the solution.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
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distilabel_metadata.role: system
distilabel_metadata.content: # 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.
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document is a discussion thread about a math problem involving optimization. The problem is to find the value of \( x \) that maximizes profit given the cost function \( C(x) = 680 + 4x + 0.01x^2 \) and the demand function \( p(x) = 12 - \frac{x}{500} \). The thread includes a question, multiple attempts at solving the problem, and a discussion about the correctness of the solution.
  
  Label: question-and-answer