$begingroup$
A boy in order to tidy his room asks his parents for a cardboard box to store lots of small toys. Unfortunately they didn't find any but only a raw cardboard sheet of dimensions 60cm x 80cm. Being very busy they told him to make one by himself.
What are dimensions of the biggest cuboid box volume the boy can make from given cardboard sheet?
Let assume that all adjacent faces which are not attached together naturally, should be glued together by strips of width at least 1cm
(There is no requirement that the strip must be attached to "source" face but you will need then another 1cm to glue it to the "source" face)
Clarification: all adjacent faces should be eventually joined together somehow (including the lid)
There is no particular requirement for the shape of joining strips. Let assume some fair minimum: a trapezoid of one side: face side length, the opposite side at least half of it and height 1cm
geometry optimization
asked 4 hours ago
mpasko256mpasko256
1,082213
$endgroup$
add a comment |
$begingroup$
A boy in order to tidy his room asks his parents for a cardboard box to store lots of small toys. Unfortunately they didn't find any but only a raw cardboard sheet of dimensions 60cm x 80cm. Being very busy they told him to make one by himself.
What are dimensions of the biggest cuboid box volume the boy can make from given cardboard sheet?
Let assume that all adjacent faces which are not attached together naturally, should be glued together by strips of width at least 1cm
(There is no requirement that the strip must be attached to "source" face but you will need then another 1cm to glue it to the "source" face)
Clarification: all adjacent faces should be eventually joined together somehow (including the lid)
There is no particular requirement for the shape of joining strips. Let assume some fair minimum: a trapezoid of one side: face side length, the opposite side at least half of it and height 1cm
geometry optimization
asked 4 hours ago
mpasko256mpasko256
1,082213
$endgroup$
$begingroup$
Is there a requirement on the angles on the edges of the joining strips? It seems like it might be a possibility to make them diagonal, but that might require particular angles.
$endgroup$
– Vicky
3 hours ago
$begingroup$
Do we assume that the joining tabs have 45 degree angles at the corners? Or are other angles allowed (like very acute angles). EDIT - I see that @Vicky asked the same thing.
$endgroup$
– chasly from UK
3 hours ago
$begingroup$
@chaslyfromUK sorry, I deleted my comment because I misunderstood your question. Please see clarification.
$endgroup$
– mpasko256
3 hours ago
add a comment |
$begingroup$
A boy in order to tidy his room asks his parents for a cardboard box to store lots of small toys. Unfortunately they didn't find any but only a raw cardboard sheet of dimensions 60cm x 80cm. Being very busy they told him to make one by himself.
What are dimensions of the biggest cuboid box volume the boy can make from given cardboard sheet?
Let assume that all adjacent faces which are not attached together naturally, should be glued together by strips of width at least 1cm
(There is no requirement that the strip must be attached to "source" face but you will need then another 1cm to glue it to the "source" face)
Clarification: all adjacent faces should be eventually joined together somehow (including the lid)
There is no particular requirement for the shape of joining strips. Let assume some fair minimum: a trapezoid of one side: face side length, the opposite side at least half of it and height 1cm
geometry optimization
asked 4 hours ago
mpasko256mpasko256
1,082213
$endgroup$
A boy in order to tidy his room asks his parents for a cardboard box to store lots of small toys. Unfortunately they didn't find any but only a raw cardboard sheet of dimensions 60cm x 80cm. Being very busy they told him to make one by himself.
What are dimensions of the biggest cuboid box volume the boy can make from given cardboard sheet?
Let assume that all adjacent faces which are not attached together naturally, should be glued together by strips of width at least 1cm
(There is no requirement that the strip must be attached to "source" face but you will need then another 1cm to glue it to the "source" face)
Clarification: all adjacent faces should be eventually joined together somehow (including the lid)
There is no particular requirement for the shape of joining strips. Let assume some fair minimum: a trapezoid of one side: face side length, the opposite side at least half of it and height 1cm
geometry optimization
geometry optimization
asked 4 hours ago
mpasko256mpasko256
1,082213
asked 4 hours ago
mpasko256mpasko256
1,082213
edited 3 hours ago
mpasko256
asked 4 hours ago
mpasko256mpasko256
1,082213
asked 4 hours ago
mpasko256mpasko256
1,082213
asked 4 hours ago
mpasko256mpasko256
1,082213
1,082213
$begingroup$
Is there a requirement on the angles on the edges of the joining strips? It seems like it might be a possibility to make them diagonal, but that might require particular angles.
$endgroup$
– Vicky
3 hours ago
$begingroup$
Do we assume that the joining tabs have 45 degree angles at the corners? Or are other angles allowed (like very acute angles). EDIT - I see that @Vicky asked the same thing.
$endgroup$
– chasly from UK
3 hours ago
$begingroup$
@chaslyfromUK sorry, I deleted my comment because I misunderstood your question. Please see clarification.
$endgroup$
– mpasko256
3 hours ago
add a comment |
$begingroup$
Is there a requirement on the angles on the edges of the joining strips? It seems like it might be a possibility to make them diagonal, but that might require particular angles.
$endgroup$
– Vicky
3 hours ago
$begingroup$
Do we assume that the joining tabs have 45 degree angles at the corners? Or are other angles allowed (like very acute angles). EDIT - I see that @Vicky asked the same thing.
$endgroup$
– chasly from UK
3 hours ago
$begingroup$
@chaslyfromUK sorry, I deleted my comment because I misunderstood your question. Please see clarification.
$endgroup$
– mpasko256
3 hours ago
$begingroup$
Is there a requirement on the angles on the edges of the joining strips? It seems like it might be a possibility to make them diagonal, but that might require particular angles.
$endgroup$
– Vicky
3 hours ago
$begingroup$
Is there a requirement on the angles on the edges of the joining strips? It seems like it might be a possibility to make them diagonal, but that might require particular angles.
$endgroup$
– Vicky
3 hours ago
$begingroup$
Do we assume that the joining tabs have 45 degree angles at the corners? Or are other angles allowed (like very acute angles). EDIT - I see that @Vicky asked the same thing.
$endgroup$
– chasly from UK
3 hours ago
$begingroup$
Do we assume that the joining tabs have 45 degree angles at the corners? Or are other angles allowed (like very acute angles). EDIT - I see that @Vicky asked the same thing.
$endgroup$
– chasly from UK
3 hours ago
$begingroup$
@chaslyfromUK sorry, I deleted my comment because I misunderstood your question. Please see clarification.
$endgroup$
– mpasko256
3 hours ago
$begingroup$
@chaslyfromUK sorry, I deleted my comment because I misunderstood your question. Please see clarification.
$endgroup$
– mpasko256
3 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Disclaimer: This is not the optimized answer but it shows how to solve it:
Since we need some strips, In order to maximize it, you need to have
6 squares taken out from the 60x80 box shown
as below;
with the dimension
$80/3$
So we create our strips out of the rest of the pieces (6.66x80) (enough to make (strip dimensions will be (fourteen of 80/3 x 1))as a result our box volume becomes ;
$(80/3)^3=18,963$.
This could be improved a bit more by increasing one edge of the 4 squares dimensions into 4 rectangles but I leave that to someone else for now :)
answered 3 hours ago
OrayOray
16k436156
$endgroup$
$begingroup$
Yes! You are certainly going in the correct way! Congratulations! :)
$endgroup$
– mpasko256
3 hours ago
add a comment |
$begingroup$
My solution:
The box is $31 times 27 times 24.5 = 20506.5 space cm^3$
The cardboard cuts out to 4 pieces, the dashed lines are folds.
The two large pieces each fold to a form a U shape.
One of them has six 1 cm tabs, the other has no tabs.
There are two separate tabs, each 2cm wide.
These two tabs are slightly longer than needed, can be cut from 31 to 27.
So the only wastage is a small area $4 times 4 = 16 space cm^2$
answered 1 hour ago
Weather VaneWeather Vane
1,64219
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Disclaimer: This is not the optimized answer but it shows how to solve it:
Since we need some strips, In order to maximize it, you need to have
6 squares taken out from the 60x80 box shown
as below;
with the dimension
$80/3$
So we create our strips out of the rest of the pieces (6.66x80) (enough to make (strip dimensions will be (fourteen of 80/3 x 1))as a result our box volume becomes ;
$(80/3)^3=18,963$.
This could be improved a bit more by increasing one edge of the 4 squares dimensions into 4 rectangles but I leave that to someone else for now :)
answered 3 hours ago
OrayOray
16k436156
$endgroup$
$begingroup$
Yes! You are certainly going in the correct way! Congratulations! :)
$endgroup$
– mpasko256
3 hours ago
add a comment |
$begingroup$
Disclaimer: This is not the optimized answer but it shows how to solve it:
Since we need some strips, In order to maximize it, you need to have
6 squares taken out from the 60x80 box shown
as below;
with the dimension
$80/3$
So we create our strips out of the rest of the pieces (6.66x80) (enough to make (strip dimensions will be (fourteen of 80/3 x 1))as a result our box volume becomes ;
$(80/3)^3=18,963$.
This could be improved a bit more by increasing one edge of the 4 squares dimensions into 4 rectangles but I leave that to someone else for now :)
answered 3 hours ago
OrayOray
16k436156
$endgroup$
$begingroup$
Yes! You are certainly going in the correct way! Congratulations! :)
$endgroup$
– mpasko256
3 hours ago
add a comment |
$begingroup$
Disclaimer: This is not the optimized answer but it shows how to solve it:
Since we need some strips, In order to maximize it, you need to have
6 squares taken out from the 60x80 box shown
as below;
with the dimension
$80/3$
So we create our strips out of the rest of the pieces (6.66x80) (enough to make (strip dimensions will be (fourteen of 80/3 x 1))as a result our box volume becomes ;
$(80/3)^3=18,963$.
This could be improved a bit more by increasing one edge of the 4 squares dimensions into 4 rectangles but I leave that to someone else for now :)
answered 3 hours ago
OrayOray
16k436156
$endgroup$
Disclaimer: This is not the optimized answer but it shows how to solve it:
Since we need some strips, In order to maximize it, you need to have
6 squares taken out from the 60x80 box shown
as below;
with the dimension
$80/3$
So we create our strips out of the rest of the pieces (6.66x80) (enough to make (strip dimensions will be (fourteen of 80/3 x 1))as a result our box volume becomes ;
$(80/3)^3=18,963$.
This could be improved a bit more by increasing one edge of the 4 squares dimensions into 4 rectangles but I leave that to someone else for now :)
answered 3 hours ago
OrayOray
16k436156
answered 3 hours ago
OrayOray
16k436156
answered 3 hours ago
OrayOray
16k436156
answered 3 hours ago
OrayOray
16k436156
16k436156
$begingroup$
Yes! You are certainly going in the correct way! Congratulations! :)
$endgroup$
– mpasko256
3 hours ago
add a comment |
$begingroup$
Yes! You are certainly going in the correct way! Congratulations! :)
$endgroup$
– mpasko256
3 hours ago
$begingroup$
Yes! You are certainly going in the correct way! Congratulations! :)
$endgroup$
– mpasko256
3 hours ago
$begingroup$
Yes! You are certainly going in the correct way! Congratulations! :)
$endgroup$
– mpasko256
3 hours ago
add a comment |
$begingroup$
My solution:
The box is $31 times 27 times 24.5 = 20506.5 space cm^3$
The cardboard cuts out to 4 pieces, the dashed lines are folds.
The two large pieces each fold to a form a U shape.
One of them has six 1 cm tabs, the other has no tabs.
There are two separate tabs, each 2cm wide.
These two tabs are slightly longer than needed, can be cut from 31 to 27.
So the only wastage is a small area $4 times 4 = 16 space cm^2$
answered 1 hour ago
Weather VaneWeather Vane
1,64219
$endgroup$
add a comment |
$begingroup$
My solution:
The box is $31 times 27 times 24.5 = 20506.5 space cm^3$
The cardboard cuts out to 4 pieces, the dashed lines are folds.
The two large pieces each fold to a form a U shape.
One of them has six 1 cm tabs, the other has no tabs.
There are two separate tabs, each 2cm wide.
These two tabs are slightly longer than needed, can be cut from 31 to 27.
So the only wastage is a small area $4 times 4 = 16 space cm^2$
answered 1 hour ago
Weather VaneWeather Vane
1,64219
$endgroup$
add a comment |
$begingroup$
My solution:
The box is $31 times 27 times 24.5 = 20506.5 space cm^3$
The cardboard cuts out to 4 pieces, the dashed lines are folds.
The two large pieces each fold to a form a U shape.
One of them has six 1 cm tabs, the other has no tabs.
There are two separate tabs, each 2cm wide.
These two tabs are slightly longer than needed, can be cut from 31 to 27.
So the only wastage is a small area $4 times 4 = 16 space cm^2$
answered 1 hour ago
Weather VaneWeather Vane
1,64219
$endgroup$
My solution:
The box is $31 times 27 times 24.5 = 20506.5 space cm^3$
The cardboard cuts out to 4 pieces, the dashed lines are folds.
The two large pieces each fold to a form a U shape.
One of them has six 1 cm tabs, the other has no tabs.
There are two separate tabs, each 2cm wide.
These two tabs are slightly longer than needed, can be cut from 31 to 27.
So the only wastage is a small area $4 times 4 = 16 space cm^2$
answered 1 hour ago
Weather VaneWeather Vane
1,64219
edited 1 hour ago
answered 1 hour ago
Weather VaneWeather Vane
1,64219
answered 1 hour ago
Weather VaneWeather Vane
1,64219
answered 1 hour ago
Weather VaneWeather Vane
1,64219
1,64219
add a comment |
add a comment |
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$begingroup$
Is there a requirement on the angles on the edges of the joining strips? It seems like it might be a possibility to make them diagonal, but that might require particular angles.
$endgroup$
– Vicky
3 hours ago
$begingroup$
Do we assume that the joining tabs have 45 degree angles at the corners? Or are other angles allowed (like very acute angles). EDIT - I see that @Vicky asked the same thing.
$endgroup$
– chasly from UK
3 hours ago
$begingroup$
@chaslyfromUK sorry, I deleted my comment because I misunderstood your question. Please see clarification.
$endgroup$
– mpasko256
3 hours ago