
MarbleDuck

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21 Feb 2013 03:40 #1
by MarbleDuck
So I have a problem that I'm not sure how to go about solving. The goal is to find the largest cubic box that can fit through a parabolic arch that is 58 ft high, and 28 ft wide.
I've figured that the equation of the parabola (with its vertex at the origin) is y=58/196x^2
but I don't know how to go about figuring out the largest square that can fit within it
I'd really appreciate any help.
youtube.com/marbleduck
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Jeff


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21 Feb 2013 05:36 #2
by Jeff
I'd help you out if i could MD, but the only thing that pops in my head when I see parabolas is finding the area under a curve......thanks a lot Calc 1 >_>
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whirligig


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21 Feb 2013 05:45 #3
by whirligig
*makes a sudden, unexpected appearance*
I'm pretty sure the diagram you drew of the box's position is correct. First, put the origin at the base of the arch, not at the vertex. This makes the equation y = 5858/196*x^2. Label one corner of the box at its upper right. Call this point (x,y). We know that y = 2x from the shape of the box (the height is twice half of the bottom side length), and y = 5858/196*x^2 from the arch (the corner touches the arch). Combine these equations to get 58/196*x^2+2x58 = 0. Solve for x, which I trust you know how to do at this point. The positive solution, x = 14/29*(âˆš890  7), is half the side length of the box.
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MarbleDuck

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21 Feb 2013 16:32 #4
by MarbleDuck
ahhhhhh okay. I was sorta getting the idea of what I had to do before, but this helped me a lot
Appreciate the help, whirli and Jeff
youtube.com/marbleduck
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21 Feb 2013 23:06 #5
by Jeff
Feb 21, 2013, 8:32am, marbleduck wrote:ahhhhhh okay. I was sorta getting the idea of what I had to do before, but this helped me a lot
Appreciate the help, whirli and Jeff
What did I do besides complaining about all I can think about with parabolas is finding the area between them and the X axis?
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MarbleDuck

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22 Feb 2013 19:06 #6
by MarbleDuck
Hey whirli
x = 14/29*(sqrt(890)  7) or x = 14/29*(sqrt(8907))
I'm terrible at these quadratic things. I always get lost among the numbers :/
...and I'm quadratic formulaing it, but I keep getting imaginary answers
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24 Feb 2013 00:00 #7
by whirligig
The 7 is outside the square root, obviously. Otherwise it wouldn't be simplified.
If you're getting nonreal answers ... my guess would be that you aren't accounting for c being negative?
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25 Feb 2013 19:35 #8
by MarbleDuck
I get confus with negatives sometimes.
So I'm still confused as to why y=2x
youtube.com/marbleduck
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