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mathcalculus

  • one year ago

FIND: The minimal average cost (ATTACHED.)

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  1. mathcalculus
    • one year ago
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  2. satellite73
    • one year ago
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    minimum is at the vertex compute \(-\frac{b}{2a}\) with \(b=700,a=1\)

  3. satellite73
    • one year ago
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    oh, i see you did that. hmmmm

  4. satellite73
    • one year ago
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    weird i guess to minimize the cost, produce nothing

  5. mathcalculus
    • one year ago
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    ?

  6. mathcalculus
    • one year ago
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    i understood part d.... it's 280

  7. satellite73
    • one year ago
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    you found the vertex correctly, but you can't produce -350 items

  8. satellite73
    • one year ago
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    where on earth did the 280 come from?

  9. mathcalculus
    • one year ago
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    That's the production level that will minimize the average cost

  10. satellite73
    • one year ago
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    oh, i guess i have no idea what an average cost is

  11. DebbieG
    • one year ago
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    YOu have cost, to find average cost divide that by x. THEN take the derivative, that is marginal average cost. Minimize THAT.

  12. satellite73
    • one year ago
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    oooooh!

  13. mathcalculus
    • one year ago
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    i did c(X)/x, differentiated, then set it to zero. and got x=280

  14. mathcalculus
    • one year ago
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    how @DebbieG

  15. satellite73
    • one year ago
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    why is that an "average cost"?

  16. mathcalculus
    • one year ago
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    The minimal average cost?

  17. DebbieG
    • one year ago
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    Because C(x) gives the total cost of producing x items. So C(x)/x gives the average cost per item, at the production level x.

  18. mathcalculus
    • one year ago
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    okaay but let's find part e please.

  19. satellite73
    • one year ago
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    learn something new every day

  20. DebbieG
    • one year ago
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    Sorry - I didn't mean to minimize the derivative... lol... I meant to minimize the average cost. Which you can do by setting the derivative of it =0. :)

  21. mathcalculus
    • one year ago
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    that's my main concern...

  22. DebbieG
    • one year ago
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    So you have average cost: A(x)=78400/x+700+x Take that derivative, set it = 0, and that's where your average cost is minimized.

  23. mathcalculus
    • one year ago
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    i did

  24. mathcalculus
    • one year ago
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    x=280

  25. mathcalculus
    • one year ago
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    i don't understand

  26. DebbieG
    • one year ago
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    OK, you good for e now?

  27. mathcalculus
    • one year ago
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    yes

  28. DebbieG
    • one year ago
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    Sorry - don't understand what?

  29. mathcalculus
    • one year ago
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    but it's wrong.

  30. mathcalculus
    • one year ago
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    d) The production level that will minimize the average cost = 280. which is correct. e) The minimal average cost= ???

  31. DebbieG
    • one year ago
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    What did you get?

  32. DebbieG
    • one year ago
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    Just evaluate the average cost function at x=280

  33. mathcalculus
    • one year ago
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    so plug x into ?

  34. DebbieG
    • one year ago
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    The average cost function: A(x)=78400/x+700+x which is just C(x)/x

  35. DebbieG
    • one year ago
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    what did you get? :)

  36. mathcalculus
    • one year ago
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    right.280!

  37. mathcalculus
    • one year ago
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    well -78400/x^2+1...

  38. mathcalculus
    • one year ago
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    then set it to zero right? @DebbieG

  39. DebbieG
    • one year ago
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    wha? no.... you found the production level that minimizes average cost already, by setting the derivative of average cost = 0, right? Now you just need to know what that average cost is - what is the average cost at that production level of x=280 So PLUG x=280 INTO the average cost function.

  40. mathcalculus
    • one year ago
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    oooooooo

  41. mathcalculus
    • one year ago
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    how about if i want to find the production level that will maximize profit.

  42. mathcalculus
    • one year ago
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    @DebbieG

  43. DebbieG
    • one year ago
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    Then you need the profit function. Then find where it is maximized, by taking its derivative and set it = 0.

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