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Math Help - Function construction.

  1. #1
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    Function construction.

    Hey folks, I have the following question, and I was hoping one of you pros could check my answer against yours or let me know if there is a better answer, mine is pretty convoluted and bulky.

    The question is;

    A salesman has a base salary of $28,000.00, and earns in addition a 5% commission per unit for the first 100 units sold, 7.5% for the next 100 units sold, and 12% for any additional units sold (i.e. above 200).

    a. If the market price per unit is $75.00, give the function of the income earned in terms of units sold.
    Now I put together the following, and it feels way to bulky, not sure if I'm making this more complicated than it is.

    f(u)=28000+(0.05)(75)u+(0.025)(75)(u-100)+(0.045)(75)(u-200)

    What do you guys think? Thanks!
    Kasper
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  2. #2
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    Smile

    hi
    i agree with what you wrote,your function is correct for any u\geq 100,but you still need confirmation
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  3. #3
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    Quote Originally Posted by Kasper View Post
    Hey folks, I have the following question, and I was hoping one of you pros could check my answer against yours or let me know if there is a better answer, mine is pretty convoluted and bulky.

    The question is;

    Now I put together the following, and it feels way to bulky, not sure if I'm making this more complicated than it is.

    f(u)=28000+(0.05)(75)u+(0.025)(75)(u-100)+(0.045)(75)(u-200)

    What do you guys think? Thanks!
    Kasper
    <br />
f\left( u \right) = \left\{ \begin{gathered}<br />
  28000 + 0.05(75)u,{\text{     if   }}0 < u \leqslant 100 \hfill \\<br />
  28000 + 0.05(75)100 + 0.075(75)(u - 100),{\text{   if   }}100 < u \leqslant 200 \hfill \\<br />
  28000 + 0.05(75)100 + 0.075(75)100 + 0.12(75)(u - 200),{\text{   if   }}u > 200 \hfill \\ <br />
\end{gathered}  \right.<br />

    <br />
 \Rightarrow f\left( u \right) = \left\{ \begin{gathered}<br />
  28000 + 3.75u,{\text{     if   }}0 < u \leqslant 100 \hfill \\<br />
  28375 + 5.625(u - 100),{\text{   if   }}100 < u \leqslant 200 \hfill \\<br />
  28937.50 + 9(u - 200),{\text{   if   }}u > 200 \hfill \\ <br />
\end{gathered}  \right.

    does that seems good?
    Last edited by Shyam; June 22nd 2009 at 05:02 PM.
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  4. #4
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    Quote Originally Posted by Shyam View Post
    <br />
f\left( u \right) = \left\{ \begin{gathered}<br />
  28000 + 0.05(75)u,{\text{     if   }}0 < u \leqslant 100 \hfill \\<br />
  28000 + 0.05(75)100 + 0.075(75)(u - 100),{\text{   if   }}100 < u \leqslant 200 \hfill \\<br />
  28000 + 0.05(75)100 + 0.075(75)100 + 0.12(75)(u - 200),{\text{   if   }}u > 200 \hfill \\ <br />
\end{gathered}  \right.<br />

    <br />
 \Rightarrow f\left( u \right) = \left\{ \begin{gathered}<br />
  28000 + 3.75u,{\text{     if   }}0 < u \leqslant 100 \hfill \\<br />
  28375 + 5.625(u - 100),{\text{   if   }}100 < u \leqslant 200 \hfill \\<br />
  28937.50 + 9(u - 200),{\text{   if   }}u > 200 \hfill \\ <br />
\end{gathered}  \right.
    i thought 28000 will remain the same,i don't get it
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