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Have "box A" (U-Haul shipping container) with dimensions length 95 inches, width 56 inches, height 83.5 inches and "box B" (couch) with dimensions length 96 inches, width 50 inches, height 34 inches. Will "box B" fit in "box A"?

It has been a long time since I took any math class, but I figured the length of the couch can easily fit within the length of the container's diagonal. I am less sure about whether the angle at which the couch would need to be tilted for its length to fit would be possible given its height.

Plan on building a scale model if this proves too uninteresting for answers. Will post pics regardless.

MJ mj
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    not a good bet. Turning the couch so that its 50 inches matches up with the 56 inches of the box, then rotating $41^\circ$ so that the far corner matches the 95 inches, I get that the dimension you want to be smaller than 83.5 comes out about 89. I like your idea of building models, though. – Will Jagy Dec 05 '18 at 18:58
  • Well, the optimal angle to just fit the 95 constraint is $$ \arccos \frac{95}{\sqrt {10372}} + \arccos \frac{96}{\sqrt {10372}} \approx 21.12^\circ + 19.50^\circ \approx 40.62^\circ $$ – Will Jagy Dec 05 '18 at 19:07
  • Bummer. Best couch in the world. – MJ mj Dec 05 '18 at 19:39
  • Unless the box is constructed of heavy material it won't support the imbalanced weight for the shipping duration anyway. – fleablood Dec 05 '18 at 19:45
  • Box A is a literal shipping container, so I think it would be fine. I'd worry more about the couch unless it were wedged in, preventing it from being knocked around. Also, the lack of balance may/may not cause issues with loading or unloading it from the train/ship. My guess is I'll have to rent a truck for the couch. – MJ mj Dec 05 '18 at 20:40
  • It's still not clear to me, (looking at these diagrams), how you started on the problem. I'll assume it's math I've long forgotten or never learned. I last took geometry in 9th grade, many years ago. – MJ mj Dec 05 '18 at 20:45
  • The way I thought about it was that width did not matter, (the couch being narrower than the container), so the thing I needed to figure out is what angle of tilt is required to fit the couch length within the container length, and whether at that angle the height of the couch versus the container height. That's where I tap out. – MJ mj Dec 05 '18 at 20:50
  • Note that the exact shape of the ends (the 50"×34" faces) might change the answer. Unfortunately, I now have the ".. in a box" song earworm, and cannot do math until I get rid of it. – Nominal Animal Dec 05 '18 at 22:30
  • @MJmj note that you should begin a comment with an @ sign and the username of the person you wish to address. That way, even though they are looking at other questions, there will be a little flag on the top of this page giving a link to your comment. I recommend you start with a 2-dimensional model, draw a rectangle 95 by 83.5 millimeters, then carefully cut out a separate rectangle 34 by 96 millimeters, and see what happens as you rotate the couch rectangle. If you have a protractor it will help. – Will Jagy Dec 05 '18 at 23:57

2 Answers2

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enter image description here

So we want $x + \frac {96}{34}\sqrt{34^2 - x^2} =95$ and that $\frac {96}{34}x + \sqrt{34^2 - x^2} < 80$

To solve:

$x + \frac {96}{34}\sqrt{34^2 - x^2} =95$

$\frac {48}{17}\sqrt{34^2 - x^2} = 95-x$

$\frac {48^2}{17^2}(34^2 - x^2) = (95-x)^2= 95^2 - 190x + x^2$

$-(\frac {48^2}{17^2}+1)x^2 + 190x + (\frac {48^2*34^2}{17^2} - 95^2)=0$

Use the quadratic formula to solve and plug it in to see if

$\frac {96}{34}x + \sqrt{34^2 - x^2} < 80$

fleablood
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well, I don't think it can be done... if you made it a full three dimensional problem and made a realistic model of the couch, it might come up possible but in a way that damages the couch

enter image description here

Will Jagy
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