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anonymous

  • one year ago

) A hotel finds that its revenue is given by R = 8000 + 760x - 30x2 when it charges 80 + 10x dollars for a room. To the nearest dollar, what is the maximum revenue it can earn?

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  1. Jack1
    • one year ago
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    \(\large R = 8000 + 760x - 30x^2\) maximum occurs at the turning point of this line can you work out the turning point of a parabola?

  2. anonymous
    • one year ago
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    i tried using the -b/2(a) and then plugging that result for x but thats where i get lost

  3. Jack1
    • one year ago
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    have you worked with derivatives yet?

  4. anonymous
    • one year ago
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    no.

  5. Jack1
    • one year ago
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    damn... k hang on

  6. Jack1
    • one year ago
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    For \(\Large y = ax^2 + bx + c\) \(\Large (c - \frac{b^2}{4a}) = ymax~~ or~~ ymin\) value at the max point

  7. Jack1
    • one year ago
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    so ur equation is \(\large R = 8000 + 760x - 30x^2\) \(\large y = ax^2 + bx + c\) so a = -30 b = 760 c = 8000 no can you solve for ymax?

  8. anonymous
    • one year ago
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    this was the answer in the help video but i think it's wrong. R=(80+10x)(100-3x) 8000-24x+1000x-30x^2 R=-30x^2+976+8000 X= -b/2a= -976/2(-30) and with all of that i was suppposed to end up with $12813 for the maximum

  9. anonymous
    • one year ago
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    the answer i got was 11400

  10. anonymous
    • one year ago
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    these were the options A) $9,920 B) $13,530 C) $12,813 D) no maximum

  11. Jack1
    • one year ago
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    hmm... 12813 is correct using ymax = c - (b^2)/4a ymax = 8000 - 760^2/4(-30) ymax = 8000 - 57760/-120 ymax = 8000 + 4813.33 = $12813 ... does this kinda make ssense tho?

  12. anonymous
    • one year ago
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    yes, i have just never used that formula before that's why. thank you :)

  13. Jack1
    • one year ago
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    welcomes ;)

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