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Homework Help: Funky function question

  1. Jul 8, 2007 #1
    1. The problem statement, all variables and given/known data

    Graph: R(x)=x(50-2x). Determine the maximum value of R.

    2. Relevant equations

    this particular question doesn't lend itself to a relevant example in the Rockswold text.

    3. The attempt at a solution

    at first, I tried graphing it on the TI as -2x^2+50x but it is a mother to try to display correctly. At that point I figured I was doing something out of proper sequence. Also, in this instance are they talking about maximum value as a component of a negative function, i.e. parabola opening downward? If so, what is the maximum value?

    Thanks in advance...
  2. jcsd
  3. Jul 8, 2007 #2
    I suppose you don't know calculus? This being the noncalculus forum. Taking the derivitive and setting it equal to zero is the classic way to solving this problem. However you can kinda do the same thing with just your wits.

    Look at the problem. Divide it into two parts. One forcing it upward and one forcing it downward. Does one part increase faster than the other? Which one is it? At which point are the two equal? What happens when they are equal?

    Answer these questions here and if you haven't got an answer by then, someone else will help you I am sure.
  4. Jul 8, 2007 #3
    Where does the function have zeros? What type of symmetry exists with a parabola? What can you say about where a parabola is maximum in relation to its zeros?
  5. Jul 8, 2007 #4
    I had to post once more... my last post count was scary.
  6. Jul 8, 2007 #5


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    Complete the square. R(x)=x(50-2x)= -2x2+ 50x= -2(x2- 25x).

    You can complete the square to write that as R(x)= -2(x- a)2+ b. If x= a then R= b while if x is any other number, R(x) is b minus something. Once you have the form R(x)= -2(x-a)2+ b, it is obvioius that the maximum value of R is b.
  7. Jul 8, 2007 #6

    Gib Z

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    Or you could find it's axis of symmetry, [itex]x= \frac{-b}{2a}[/itex]
  8. Jul 8, 2007 #7
    OK, how about the NON-math/accounting/physics major's interpretation of the above?:redface:

    I'm looking at this and really am not sure of where to complete the square. Would I be halving 50 and then squaring the result, or does another value represent b?

    A fresh morning look at the problem assumes it to be something to the effect of x^2 - 25x + (-25/2)^2= 0+(-25/2)^2. This gives me


    which would leave us with something to the effect of


    and then

    x-12.5=+/- V'156.25

    which I find to be


    and then -12.5+12.5=0

    while 12.5+12.5=25

    so x=0 and x=25

    making max value of R=25?

    could this be correct? and I would graph this... how?
    Last edited: Jul 8, 2007
  9. Jul 8, 2007 #8

    Math Is Hard

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    Maybe this will help (to get started, anyway):There are two places that the height of your curve, R(x), is going to be zero: when x = 0 and when x = 25. Those are the points where it crosses the x-axis. Draw your x and y axes and label the x-axis with a point at zero on the line and a point at 25. Now, draw an arc swooping up from the point (0,0) and then back down to the point (25,0).

    p.s. There's a nice little tutorial on the topic here:
    Last edited: Jul 8, 2007
  10. Jul 8, 2007 #9

    Okay we can show from quadratic of the form y = ax^2+bx+c that we can derive the parbolic form y-k=a(x-h)^2 ( where (h,k) is the vertex of the parbola ) by completing the square on the quadratic form.

    Doing that we get y-(-b^2 - 4ac/2a) = a(x-(-b/2a)^2, well it is easy to see just by comparsion what h,k are ( which we can say are x,y ) here x= -b/2a and y= -b-4ac/4a. So that was what Gibz was saying.

    R(x)=(50-2x), Let R(x) = y factoring we get y= (-2x+50) = -2(x^2 -25x)

    I think i should say that the value of a tells us whether the function is max or min if the vaue of a is greater that 0 the function has a min value if is is less that 0 then we have ourselves a max value, as you can see
    -2 is less than 0so we have a mxaimum value.

    you are very close, try completing by changing R(x) to y, completing the square and comparing the value you get to what i showed about the vertex of a parbola.

    Last edited: Jul 8, 2007
  11. Jul 8, 2007 #10


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    Actually, for a "NON-math/accounting/physics major" you've done very well. But when you said
    You are solving for x, not R! What you found were the values of x when R= 0, not the maximum value of R.

    Yes, (x- a)2= x2- 2ax+ a2. If you want that to be the same as x2- 25x+ ..., then 2a must be 25 and so a must be 12.5. Of course then a2= 12.52= 156.25.
    x2- 12x= x2- 12x+ 156.25- 156.25= (x-12.5)2- 156.25.

    HOWEVER, your original problem was R(x)= x(50- 2x)= -2x2+ 50x. If you divide both sides by -2, you get (-1/2)R(x)= x2- 25x= (x-12.5)2- 156.25. Now multiply both sides by -2 to get back to f(x):
    R(x)= -2(x-12.5)2+ 312.5

    When x= 12.5, R(x) is 312.5. If x is any other number, R(x) is smaller than 312.5.
  12. Jul 8, 2007 #11
    quick question Hallofivy, please forgive my ignorance, how do get to multi quote like you did? I mean i see the button but i still dont get it.

  13. Jul 8, 2007 #12
    the moment of truth...?

    soooo... if I were to graph this beast, I would be looking at a negative quadratic function whose vertex would be some value of x, and y=312.5? Sounds like a major SWAG of a graph, to be sure...
  14. Jul 8, 2007 #13

    Math Is Hard

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    Halls gave you the x-coordinate of the vertex. Re-read the post by him you just quoted.
  15. Jul 8, 2007 #14
    Say you want to quote 3 posts: click on two of the posts multiquote you want to include, they will be highlighted in blue, then click on quote rather than multiquote in the last post you want to include and they will all be displayed in a single new post.
    Last edited: Jul 8, 2007
  16. Jul 8, 2007 #15
    OK thanks to all... I graphed the beast with x=12.5 and Rmax=312.5. Prof is a sadist, that much I know.
  17. Jul 8, 2007 #16

    Cool Thanks
  18. Jul 8, 2007 #17


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    y= 312.5- 2(x-12.5)2.

    Not just "some" x but quite obviously x= 12.5.
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