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Volume of a Box + Finding all solutions

  1. Sep 4, 2009 #1
    1. See Attachment



    2. None



    3. a. I found the volume of the box which is like this
    x(24-2x)(24-2x)
    = x(576-96x+4x2)
    = 4x3 - 96x2 + 576x
    = 4(x3 -24x2 +144x)
    but how do I find the domain?
    b. and c.How do I use my graphing calculator to graph this because when I graph I only get a line?



    1. Find all solutions of the equation.
    2cos2[tex]\Theta[/tex] - [tex]\sqrt{3}[/tex] = 0




    2. None



    3. 2cos2[tex]\Theta[/tex] - [tex]\sqrt{3}[/tex] = 0
    = 2cos2[tex]\Theta[/tex] = [tex]\sqrt{3}[/tex]
    = cos 2[tex]\Theta[/tex] = [tex]\sqrt{3}[/tex]/2
    = 2[tex]\Theta[/tex] 30, 330, 390, 690
    = [tex]\Theta[/tex] = 15, 165, 195, 345 degrees
    Did I do this right?

    Thank You!!
     

    Attached Files:

  2. jcsd
  3. Sep 4, 2009 #2

    symbolipoint

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    Knowing if your initial function properly represents the description from your exercise's wording is difficult or impossible. Best is give the full description of the problem, and the diagram if this was included. (I refer, "to I found the volume of the box which is like this
    x(24-2x)(24-2x). )
     
  4. Sep 4, 2009 #3

    VietDao29

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    Since you didn't provide us with the full description of the problem, it's almost impossible to check your work. So, assume that there's no error in your work. The domain in this case, is the set of x, where all sides' length are positive, i.e, to find the domain, you have to solve the following set of inequalities:

    [tex]\left\{ \begin{array}{c} x > 0 \\ 24 - 2x > 0 \end{array} \right.[/tex]

    Umm, I'm not sure what you mean, how can you actually get a line, if you are graphing a cubic function?

    Well, this is not "all possible solutions". Does it mention anywhere in your book that:

    [tex]\cos \theta = \alpha \Leftrightarrow \theta = \pm \arccos (\alpha) + 2k\pi , \quad \mbox{where } k \in \mathbb{Z}[/tex]?

    Or, if you are working with degrees, then:
    [tex]\cos \theta = \alpha \Leftrightarrow \theta = \pm \arccos (\alpha) + k(360 ^ \circ) , \quad \mbox{where } k \in \mathbb{Z}[/tex]
     
  5. Sep 6, 2009 #4

    Mentallic

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    Follow VietDao29's advice to solve the inequalities.

    Either you typed in [tex]x(24-2x)^2=0[/tex] which will give you a vertical line at x=0 and x=12, or, your graphing calculator's range and domain was too small to show the features such as the turning points of the cubic function [tex]y=x(24-2x)^2[/tex]
    What I mean is, for the domain [tex]0<x<12[/tex] the maximum turning point is actually very large (over 1000). If your displayed range on the graph calculator is much smaller, then it will probably just look like a vertical line at x=0 (and x=12 if that domain is displayed too). Do you know what this turning point corresponds to physically?


    Yes this is correct! Assuming the question specifically stated for [tex]0^o\leq \theta \leq 360^o[/tex] or this is what is expected for your class.
     
    Last edited by a moderator: Sep 6, 2009
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