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Quick quadratic questions

  1. Oct 3, 2006 #1
    1) "find the vertex algebraically"

    y= -x^2 - 3
    y = (x - 3)(x+1)??
    (the -x is throwing me off)

    2) a movie theatre sells tickets for $8.5 each. they are considering raising the prices but know that for every 50 cents the price is raised, 20 fewer people go to the movies. R = -40c^2 + 84c
    c = ticket cost
    R = revenue

    what should the theatre charge to maximize revenue?

    for this one i can probably just use my graphing calculator to find the roots?. but what values should i put in the window? (new to using graphing calculator)

    ~Amy
     
  2. jcsd
  3. Oct 3, 2006 #2
    1. The vertex is [tex] (-\frac{b}{2a}, f(-\frac{b}{2a})) [/tex]. So [tex] y = -x^{2} - 3[/tex], [tex] b = 0 [/tex]. Therefore, the vertex is [tex] (0, f(0)) [/tex] or [tex] (0,-3) [/tex].


    2. Find the vertex of [tex] R(c). [/tex] [tex] b = 84, a = -40 [/tex].
     
    Last edited: Oct 3, 2006
  4. Oct 3, 2006 #3
    thanks!

    for 2) if i do the quadratic formula i end up with -85, and -82.95.. does this mean anything?

    ~Amy
     
  5. Oct 3, 2006 #4
    Since [tex] R(c) [/tex] is quadratic, then the maximum revenue would occur at the vertex, not at the roots. So it would be [tex] (\frac{-84}{-80}, 44.1) [/tex] or [tex] (1.05, 44.1) [/tex].
     
    Last edited: Oct 3, 2006
  6. Oct 3, 2006 #5
    thanks! where did the 46.2 come from?

    ~Amy
     
  7. Oct 3, 2006 #6
    [tex] R(1.05) = -40(1.05)^{2} + 84(1.05) = 44.1 [/tex]. Should not be 46.2. My bad.
     
  8. Oct 3, 2006 #7
    actually does that answer the question about whether or not they should raise the ticket price or not? im missing more pieces of this puzzle.

    they should increase the price to $44?

    ~Amy
     
  9. Oct 3, 2006 #8
    they should charge $1.05 according to the revenue function. Is there more information to this question?
     
  10. Oct 3, 2006 #9
    that's all the info to the question. but the answer sounds a bit iffy. thanks, but i dunno. if you calculate $1.50 for c into the original formula:
    R = -40c^2 + 84c
    the revenue = 36

    unless im calculating that wrong

    ~Amy
     
  11. Oct 3, 2006 #10
    Its $1.05 not $1.50.
     
  12. Oct 3, 2006 #11
    k, with $1.05 i calculated the revenue to be $1852.2

    with the original $8.5 i calculated it to be $116314

    ~Amy
     
  13. Oct 3, 2006 #12
    Are you sure? In [tex] R(c) = -40c^2 + 84c [/tex], the -40 is outside of the squared expression. So you square c first, then multiply it by -40 and add it to 84c.

    [tex] -40c^2 [/tex] does not equal [tex](-40c)^{2} [/tex]
     
    Last edited: Oct 3, 2006
  14. Oct 3, 2006 #13
    k, sorry for the hassle, looks like you are right :blushing:

    thanks again!

    ~Amy
     
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