Box in a box: will it fit? or whether or not I can take my couch to AK












4














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.






geometry







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  • 1




    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 5 '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 5 '18 at 19:07












  • Bummer. Best couch in the world.
    – MJ mj
    Dec 5 '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 5 '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 5 '18 at 20:40


















4














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.






geometry







share|cite|improve this question


















  • 1




    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 5 '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 5 '18 at 19:07












  • Bummer. Best couch in the world.
    – MJ mj
    Dec 5 '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 5 '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 5 '18 at 20:40
















4












4








4







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.






geometry







share|cite|improve this question













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.






geometry




geometry






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asked Dec 5 '18 at 17:59









MJ mjMJ mj

211




211








  • 1




    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 5 '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 5 '18 at 19:07












  • Bummer. Best couch in the world.
    – MJ mj
    Dec 5 '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 5 '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 5 '18 at 20:40
















  • 1




    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 5 '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 5 '18 at 19:07












  • Bummer. Best couch in the world.
    – MJ mj
    Dec 5 '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 5 '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 5 '18 at 20:40










1




1




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 5 '18 at 18:58




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 5 '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 5 '18 at 19:07






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 5 '18 at 19:07














Bummer. Best couch in the world.
– MJ mj
Dec 5 '18 at 19:39




Bummer. Best couch in the world.
– MJ mj
Dec 5 '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 5 '18 at 19:45




Unless the box is constructed of heavy material it won't support the imbalanced weight for the shipping duration anyway.
– fleablood
Dec 5 '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 5 '18 at 20:40






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 5 '18 at 20:40












2 Answers
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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



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






share|cite|improve this answer





























    0














    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






    share|cite|improve this answer























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      2 Answers
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      2 Answers
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      0














      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



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






      share|cite|improve this answer


























        0














        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



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






        share|cite|improve this answer
























          0












          0








          0






          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



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






          share|cite|improve this answer












          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



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







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 5 '18 at 19:18









          fleabloodfleablood

          68.7k22685




          68.7k22685























              0














              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






              share|cite|improve this answer




























                0














                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






                share|cite|improve this answer


























                  0












                  0








                  0






                  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






                  share|cite|improve this answer














                  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







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 5 '18 at 19:59

























                  answered Dec 5 '18 at 19:41









                  Will JagyWill Jagy

                  102k5100199




                  102k5100199






























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