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Finding the length and width of rectangle

  1. Oct 23, 2014 #1
    Mod note: Thread originally posted in a technical math section, so is missing the homework template.
    I am trying to solve this problem it states as follows "A piece of wire 24 cm long has the shape of a rectangle.
    (a) Given that the width is W cm, show that the area, A cm^2 of the rectangle is given by the function A=36-(6-W)^2. (b) Find the greatest possible domain and corresponding range of the function.

    My challenge is in part (a) this is how i attempted it
    2L+2W= 24 , L+W=12 , L=12-W , .........W(12-W)= 36-(6-W)^2...........where 12W-W^2=36-36+12W-W^2, whence 12W-W^2=12W-W^2 this is correct but is this the way to show it?:L
     
    Last edited by a moderator: Oct 23, 2014
  2. jcsd
  3. Oct 23, 2014 #2

    Mentallic

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    This part is obviously unhelpful.

    You have that

    [tex]12w-w^2=36-36+12w-w^2[/tex]

    Now you should do

    [tex]=36-(w^2-12w+36)[/tex]
    [tex]=36-(w-6)^2[/tex]
    [tex]=36-(6-w)^2[/tex]

    or if you're proficient enough at spotting binomials (you already know where you are going to end up so you don't even really need to be in this case), then just simply do

    [tex]12w-w^2=36-36+12w-w^2[/tex]
    [tex]=36-(36-12w+w^2)[/tex]
    [tex]=36-(6-w)^2[/tex]
     
  4. Oct 23, 2014 #3
    I appreciate but you have just re- expressed the rhs which we have already been given, in the question.... we are supposed to come up with the left hand side expression which should be equal to A=36-(6-W)^2 (the rhs)
     
  5. Oct 23, 2014 #4

    HallsofIvy

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    What are you referring to when you say "left hand expression"?
     
  6. Oct 23, 2014 #5
    ##12w-w^2##
     
  7. Oct 23, 2014 #6

    Mentallic

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    You did come up with the correct equation for the area in terms of the width, which is A=w(12-w). Now you just need to manipulate this expression to show that it is in fact equivalent to 36-(6-w)2. Once you've done that, you've answered the question.
     
  8. Oct 23, 2014 #7

    RUber

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    This is absolutely correct. Then you are asked to show: A= W*L = W(12-W) is equivalent to 36-(6-W)^2.
    I think you did this too. You expanded the right hand side and showed it was equal to the left hand side.

    If you wanted to start from the left hand side and manipulate it to show that you can get the right hand side, just add zero in the form of 36-36. Then regroup.
    Either way, as long as you present it clearly, you have shown that the two expressions are equivalent.
     
  9. Oct 24, 2014 #8

    vela

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    There are better ways than what you wrote. First, don't start with W(12-W) = 36-(6-W)2 because that's what you're supposed to prove. If you start off by writing that down, you're assuming what's to be shown, i.e., begging the question. Second, what you wrote means:
    \begin{align}
    W(12-W) &= 36-(6-W)^2 \Rightarrow \\
    12W-W^2 &= 36-36+12W-W^2 \Rightarrow \\
    12W-W^2 &= 12W-W^2.
    \end{align} It's not correct logically to start with some statement, reach a true conclusion, and infer that the initial statement was true. You could fix it in this case by simply pointing out each step is reversible, but I think that's kind of clumsy. In problems like this, it's better to start with one side and manipulate it into being equal to the second side, which is what Mentallic did above.
     
  10. Oct 25, 2014 #9

    HallsofIvy

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    Starting with what you want to prove and using invertible steps to arrive at an "obviously true" conclusion (say, one of your hypotheses) is called "synthetic proof" and is often used.
     
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