In class we had this problem and the answer the teacher gave does not satisfy me:
You have a 24x24 square piece of material. Your job is to cut equally sized squares out of the material's corners. The remaining material will for a net for a box. This box will not have a cover.
The teacher gave the answer of 4. However, I am willing to bet money that a cube contains has the highest volume to surface area ratio. I see three ways to resolve this contradiction. First, I am wrong and cubes are not inherently more efficient than other prism. However, that does raise the question of what is the most efficient prism. Second, something about this problem prevents cubes from being as efficient as they usually are. Third, my teacher make a mistake in his calculations.
Edit: This is embarrassing; I forgot to write what the question actually was. The question was what value for s gives the largest box, where s is the length of a side of one of the squares cut out of the corner?
Efficient Prizims
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Efficient Prizims
Last edited by jewish_scientist on Wed Apr 06, 2016 1:53 am UTC, edited 1 time in total.
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Re: Efficient Prizims
What's the question? (I'm assuming it's, what size of square gives the biggest volume?)jewish_scientist wrote:You have a 24x24 square piece of material. Your job is to cut equally sized squares out of the material's corners. The remaining material will for a net for a box. This box will not have a cover.
The teacher gave the answer of 4.
Remember, the surface area isn't constant; you're removing more surface area when you cut out bigger squares.However, I am willing to bet money that a cube contains has the highest volume to surface area ratio.
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Re: Efficient Prizims
Yeah, you would need to cut out four 4x4 squares to maximize the volume of the box in this case. Like chridd said, the surface area decreases more if you cut out larger squares, so you want enough "box" left over. Also, it's important that this box is not closed. A closed cube may have the best ratio of volume to surface area, but cube with one face missing does not. You can afford to make the open end a bit bigger than usual because those two dimensions contribute less to the total surface area than the height does.
You really need the tools of calculus to solve for the largest possible volume. If you cut four squares each with side length x, then you can write down the volume of the box in terms of x. Then finding the maximum is relatively straightforward if you know how do take derivatives.
You really need the tools of calculus to solve for the largest possible volume. If you cut four squares each with side length x, then you can write down the volume of the box in terms of x. Then finding the maximum is relatively straightforward if you know how do take derivatives.
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Re: Efficient Prizims
You should be able to verify that the teacher didn't make a mistake without having to ask here. Cutting out four 4x4 squares from the corner gives you a 16x16x4 box, while cutting out four 8x8 squares gives you an 8x8x8 box. Just multiply to see which one is in fact bigger. (Proving that it's the maximum is harder, but this is a common problem given in calc classes and if you know calc you should be able to verify that as well.)
1) As chridd says, cutting out the corners lowers the surface area. If you cut out four 4x4 squares from the corners, you're left with 512 square units of material. If you cut out 8x8 squares instead (to make a cube), you only have 320 square units left.
2) Even if you had a constant amount of material to work with (as in you could somehow rearrange it to make your box without having to cut any pieces off and discard them), you don't have to cover the top. That means you're not actually worried about the whole surface area. (As it turns out the volume of an opentopped box using a given (constant) area of material is maximized when it's half a cube rather than a whole cube.)
No, you're right that cubes are the most efficient in terms of volumetototalsurfacearea ratio. However, you're failing to consider two things.jewish_scientist wrote:First, I am wrong and cubes are not inherently more efficient than other prism.
1) As chridd says, cutting out the corners lowers the surface area. If you cut out four 4x4 squares from the corners, you're left with 512 square units of material. If you cut out 8x8 squares instead (to make a cube), you only have 320 square units left.
2) Even if you had a constant amount of material to work with (as in you could somehow rearrange it to make your box without having to cut any pieces off and discard them), you don't have to cover the top. That means you're not actually worried about the whole surface area. (As it turns out the volume of an opentopped box using a given (constant) area of material is maximized when it's half a cube rather than a whole cube.)
Re: Efficient Prizims
As gmalivuk said, this is easy to do if you know some calculus, but it's not too hard to verify that s=4 gives the maximum volume for the open box.
To make the arithmetic simpler, I'll scale the square down by ¼, so that s=1 and the original square is 6 x 6.
The base of the box has length and width 6  2s and height s, so its volume is
V = s(6  2s)²
Obviously, s has to be between 0 and 3.
To show that s = 1 gives the maximum volume we let s = 1 + d, where d is a small number.
We know that 0 < s < 3, so 0 < 1 + d < 3, that is, 1 < d < 2.
Substituting s = 1 + d into our volume formula gives us
V = (1 + d)(6  2  2d)²
= (1 + d)(4  2d)²
= (1 + d)2²(2  d)²
= 4(1 + d)(4  4d + d²)
= 4(4  4d + d² + 4d  4d² + d³)
V = 4(4  3d² + d³)
Notice how we only have d² and d³ terms in that volume formula, there's no d term. That's a symptom of being at a maximum (or minimum) value, and you'll fully appreciate why that is so when you learn differential calculus.
V = 4(4  3d² + d³)
V = 4(4  d²(3  d))
When d=0, V = 4 x 4 = 16.
For any d not equal to zero in the range 1 < d < 2 the d²(3  d) term will always be negative, since d² is always positive, and 4 > 3  d > 1, so (3  d) is also positive. So V must be at its maximum when d = 0, that is, when s = 1.
Here's how we show that 4 > 3  d > 1:
1 < d < 2
1 > d > 2
3 + 1 > 3  d > 3  2
4 > 3  d > 1
Here's a table of V for values of s near 1 (calculated using a bit of Python code).
And here's the Python code used to create that table:
That code will run on Python 2 or Python 3.
To make the arithmetic simpler, I'll scale the square down by ¼, so that s=1 and the original square is 6 x 6.
The base of the box has length and width 6  2s and height s, so its volume is
V = s(6  2s)²
Obviously, s has to be between 0 and 3.
To show that s = 1 gives the maximum volume we let s = 1 + d, where d is a small number.
We know that 0 < s < 3, so 0 < 1 + d < 3, that is, 1 < d < 2.
Substituting s = 1 + d into our volume formula gives us
V = (1 + d)(6  2  2d)²
= (1 + d)(4  2d)²
= (1 + d)2²(2  d)²
= 4(1 + d)(4  4d + d²)
= 4(4  4d + d² + 4d  4d² + d³)
V = 4(4  3d² + d³)
Notice how we only have d² and d³ terms in that volume formula, there's no d term. That's a symptom of being at a maximum (or minimum) value, and you'll fully appreciate why that is so when you learn differential calculus.
V = 4(4  3d² + d³)
V = 4(4  d²(3  d))
When d=0, V = 4 x 4 = 16.
For any d not equal to zero in the range 1 < d < 2 the d²(3  d) term will always be negative, since d² is always positive, and 4 > 3  d > 1, so (3  d) is also positive. So V must be at its maximum when d = 0, that is, when s = 1.
Here's how we show that 4 > 3  d > 1:
1 < d < 2
1 > d > 2
3 + 1 > 3  d > 3  2
4 > 3  d > 1
Here's a table of V for values of s near 1 (calculated using a bit of Python code).
Code: Select all
s V
0.5 12.5
0.6 13.824
0.7 14.812
0.8 15.488
0.9 15.876
1.0 16.0
1.1 15.884
1.2 15.552
1.3 15.028
1.4 14.336
1.5 13.5
And here's the Python code used to create that table:
Code: Select all
from __future__ import print_function
print('s V')
for i in range(5, 6):
s = 1 + 0.1 * i
V = s * (6  2*s) ** 2
print(s, V)
That code will run on Python 2 or Python 3.
Re: Efficient Prizims
Here's another way to see that s = 4 is the correct answer; this involves a bit of trickery and a smidgen of intuition.
If we cut out four sbys squares from the corners and fold up, we see that we get an sby(242s)by(242s) box. As such, its volume is
V = s(242s)^2.
We wish to maximize V. Here's the trickery: instead, let us maximize 4V (which will lead to the same dimensions, obviously, as V is maximized precisely when 4V is maximized).
4V = 4s(242s)^2 = (4s)*(242s)*(242s).
So 4V is the product of three positive real numbers whose sum is 48. And here's the smidgen of intuition: it should be obvious to you (for very similar reasons to why you wanted to claim that the cube had the best volume across prisms of a given surface area) that the maximum value of this product of three real numbers whose sum is constant occurs when the three real numbers are equal, that is, when 4s = 242s = 242s = 16, i.e. when s = 4.
So s = 4 gives the largest volume.
If we cut out four sbys squares from the corners and fold up, we see that we get an sby(242s)by(242s) box. As such, its volume is
V = s(242s)^2.
We wish to maximize V. Here's the trickery: instead, let us maximize 4V (which will lead to the same dimensions, obviously, as V is maximized precisely when 4V is maximized).
4V = 4s(242s)^2 = (4s)*(242s)*(242s).
So 4V is the product of three positive real numbers whose sum is 48. And here's the smidgen of intuition: it should be obvious to you (for very similar reasons to why you wanted to claim that the cube had the best volume across prisms of a given surface area) that the maximum value of this product of three real numbers whose sum is constant occurs when the three real numbers are equal, that is, when 4s = 242s = 242s = 16, i.e. when s = 4.
So s = 4 gives the largest volume.
(∫p^{2})(∫q^{2}) ≥ (∫pq)^{2}
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Thanks, skeptical scientist, for knowing symbols and giving them to me.
Re: Efficient Prizims
My intuition for why a halfcube is the most efficient shape for a box with no top:
Spoiler:
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Re: Efficient Prizims
16*16*4 isn't a half cube.
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Re: Efficient Prizims
Right. A half cube would be the optimal shape if you were told to make a box with a specified surface area. However because in this problem your box is built from less and less area the taller it is, it is reasonable to expect that the optimal shape is even shorter than that.
Re: Efficient Prizims
Right, I was solving the wrong problem. First coffee, then post. Sorry.
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