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A few more questions

  1. May 27, 2006 #1

    danago

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    Gold Member

    Hey. This is my second topic like this. I have exams coming up soon, and was given a few practice exams for a few of my classes. Some of the questions in the exams i am unable to do, or am unsure of my solution. All help is greatly appreciated.

    1. Solve for x and y:
    [tex]
    \begin{array}{l}
    \frac{{2^x }}{{4^y }} = 16 \\
    \log _3 (x + y) = 1 \\
    \end{array}
    [/tex]


    What i understand is that i need to make both equations in terms of either x or y, and then solve simultaneously. I managed the first one, but am a bit stuck with the second. For the first one i got:

    [tex]
    \begin{array}{l}
    \frac{{2^x }}{{4^y }} = 16 \\
    \therefore x\log 2 - y\log 4 = log 16 \\
    \therefore x\log 2 = y\log 4 + log 16 \\
    \therefore x = \frac{{y\log 4 + log 16}}{{\log 2}} \\
    \end{array}
    [/tex]

    ______________________________________________​

    2. A company sells items such that the revenue, in dollars, recieved from selling x items is given by:
    [tex]
    R(x) = 2x(40x - x^2 )
    [/tex]

    The cost, in dollars, of producing x items is given by:
    [tex]
    C(x) = 2(20x + 200)
    [/tex]

    There is a set-up cost of $1000

    a) Show clearly that the profit is given by:
    [tex]
    P(x) = 60x^2 - 2x^3 - 200x - 1000
    [/tex]
    b) Determine how many items the company needs to sell to be profitable
    c) Determine how many need to be sold for maximum profit
    d) Determine the maximum profit possible
    e) Determine the marginal revenue on 10 items. Explain clearly what this represents.

    a) I said that:
    [tex]
    \displaylines{
    P(x) = R(x) - C(x) \cr
    = (80x^2 - 2x^3 ) - (20x^2 + 200x + 1000) \cr
    = 60x^2 - 2x^3 + 200x - 1000 \cr}
    [/tex]

    b) For the company to reach a point where any further revenue is profit, P(x) would be equal to 0 (ie. [tex]P(x) = 0[/tex])

    [tex]P(x) = 0[/tex] when [tex]x=7.4[/tex], so i said that for the company to be profitable, [tex]x > 7.4[/tex].

    Aswell as the cubic equalling 0 when x=7.4, it also equals 0 when x=25.3. So should i say that for the company to be profitable, it needs to sell x number of items, where:
    [tex]25.3 > x > 7.4[/tex]
    ?

    c) For maximum profit, i just graphed the function P(x), and looked for a maximum where [tex]25.3 > x > 7.4[/tex]. The answer i got was x=18.2.

    d) For max profit, i just found P(18.2), which equals approximately $3177

    e) Im not entirely sure what this means. What is marginal revenue? I looked it up on google, and from what i understand, its the change in revenue per unit. Change in revenue would be given by:
    [tex]
    \frac{{dR}}{{dx}} = 160x - 6x^2
    [/tex]

    When 10 items are sold (ie. x=10):
    [tex]
    \frac{{dR}}{{dx}} = 1000
    [/tex]

    Does this mean that when 10 items are sold, the revenue is increasing by $1000 per item sold?

    To me, that all looks correct, but id like some confirmation. Also, for this type of question, since 1 item is considered 1 unit, instead of using decimal numbers, should i round my numbers to whole numbers. So instead of [tex]25.3 > x > 7.4[/tex], should i be saying [tex]25 > x > 8[/tex].

    ______________________________________________​

    2. c is a vector of magnitude 5 units with a bearing of 105 degrees. d is a vector of magnitude 7 units and a bearing of 320 degrees. Draw a clear diagram and then calculate the magnitude of c-d.

    Heres my diagram:
    [​IMG]

    I put them in component form...so:
    [tex]
    \begin{array}{l}
    \overrightarrow c = \left( {\begin{array}{*{20}c}
    {5\cos 15} \\
    { - 5\sin 15} \\
    \end{array}} \right) \\
    \overrightarrow d = \left( {\begin{array}{*{20}c}
    { - 7\sin 40} \\
    {7\cos 40} \\
    \end{array}} \right) \\
    \end{array}
    [/tex]

    [tex]
    \begin{array}{l}
    \overrightarrow c - \overrightarrow d = \left( {\begin{array}{*{20}c}
    {5\cos 15 + 7\sin 40} \\
    { - 5\sin 15 - 7\cos 40} \\
    \end{array}} \right) \\
    \therefore \left| {\overrightarrow c - \overrightarrow d } \right| = 11.46 \\
    \end{array}
    [/tex]

    on a bearing of 126 degrees.
     
    Last edited: May 27, 2006
  2. jcsd
  3. May 27, 2006 #2

    Integral

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    Staff Emeritus
    Science Advisor
    Gold Member

    For #1.
    You must take the log of BOTH sides of the equation. So your left hand side should be log 16. It may be interesting to use use log 2 .

    For the second equation, Consider the definition of loga:

    if x = ay then y=logax
     
  4. May 27, 2006 #3

    danago

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    Gold Member

    Oops. Just realised that with that first one, i copied my working out wrong. I did originally take logs of both sides. And yea, with the second one, it just clicked to me that a the log of the same number as its base is 1, so x+y=3.

    Thanks for the reply
     
  5. May 27, 2006 #4

    Hootenanny

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    Staff Emeritus
    Science Advisor
    Gold Member

    Your vector question is also correct. Note that the question only required the magnitude, you gave the bearing also. If this question was in an exam I would not waste time calculating the direction if it was not required; however, I see no harm doing it here or in an exam if you have extra time and are bored :wink:

    ~H
     
  6. May 27, 2006 #5

    danago

    User Avatar
    Gold Member

    Well actually, i did need to find direction aswell, i just forgot to write it in my question :P

    And thanks again for the reply.
     
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