file Math help?

  • MarbleDuck
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21 Feb 2013 03:40 #1 by MarbleDuck
Math help? was created 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.

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  • Jeff
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21 Feb 2013 05:36 #2 by Jeff
Replied by Jeff on topic Math help?
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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21 Feb 2013 05:45 #3 by whirligig
Replied by whirligig on topic Math help?
*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 = 58-58/196*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 = 58-58/196*x^2 from the arch (the corner touches the arch). Combine these equations to get 58/196*x^2+2x-58 = 0. Solve for x, which I trust you know how to do at this point. The positive solution, x = 14/29*(√890 - 7), is half the side length of the box.

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21 Feb 2013 16:32 #4 by MarbleDuck
Replied by MarbleDuck on topic Math help?
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
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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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22 Feb 2013 19:06 #6 by MarbleDuck
Replied by MarbleDuck on topic Math help?
Hey whirli

x = 14/29*(sqrt(890) - 7) or x = 14/29*(sqrt(890-7))

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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  • whirligig
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24 Feb 2013 00:00 #7 by whirligig
Replied by whirligig on topic Math help?
The 7 is outside the square root, obviously. Otherwise it wouldn't be simplified.

If you're getting non-real 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
Replied by MarbleDuck on topic Math help?
I get confus with negatives sometimes.

So I'm still confused as to why y=2x

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