The longest ladder if it were a straight line would be 5√5 and the angle it would make with the opposite wall would be 26.565 deg. With this geometry if we assume that the maximum width that will pass around the corner will be just under 1 then we could work out the maximum length of the ladder as approxly. 8.68034 and thus maximum area would be approxly. 8.68. Now someone please explain what's wrong with this argument? Ajit

Ok... so here is what I did. If you just use a line first and no shape. You will find that the maximum length that can fit through that hole is 13. How do I know this? Well to understand this, first we start the line on the very left side standing up. then, as you move the bottom of the line to the right, the top of the line will naturally move down. This happens on the x axis and the y axis. And since they both move at the same rate, the line itself follows a circular path. It follows the bottom left side of a circle. The equation of a circle is (x-k)^2 + (y-h)^2 = r^2. Since the circle is going to have the center point (r,r), we can just change the formula to (x-r)^2 + (y-r)^2 = r^2. *NOTE* r is also the length of line *NOTE* Now, we can just plug in (8,1) to find out what equation of circle falls on that point. (8-r)^2 + (1-r)^2 = r^2. You basically end up with two answers here because there are two circles with center (r,r) that have that point lying on it's path. But I found that 13 was the answer because it was the one where the point (8,1) landed on it's bottom left part.

Now, that wasn't so bad. However the rectangle part is a little different and a little more difficult. If you just think about two lines running right next to each other instead, you can see that now you just have two lines running the same paths, except one is a little above the other. In fact, think of two circles, it's just a circle inside of another circle. One line follows the path of the big circle, and the other line follows the path of the circle inside of the bigger circle. Well we need to figure out two equations of these two circles. Since they BOTH lie on the same center (r,r) *NOTE* r is the radius of the big circle *NOTE*, we can say (x-r)^2 + (y-r)^2 = r^2 for the big circle like before, and (x-r)^2 + (y-r)^2 = L^2 (L being radius of smaller circle). Now we set x and y to (8,1) and we finally have our equation! This equation is based on the bigger circle that MUST lie on the x and y axis. Now you can just play with numbers to see what you get. (r-L) is our width of the rectangle and L is the length of the rectangle. r(r-L) would be your area that you get. And, you can use the equation above (L^2 = ....) and solve for L, the plug it into r(r-L) to get ONLY r's in that function. To maximize it, you just need to take the derivative of that with respect to r and set it equal to 0. But this is rather difficult and I'm not sure how I would do it.

But on to your question about if it could "really" fit in there. I don't see why you couldn't put it in there. Unless there is some kind of physics that disallows of it?

It is a copyrighted article. You may need to purchase it online. Or your better chance is that you can borrow it from a university library. Here is another link: http://www1.american.edu/academic.depts/cas/mathstat/People/kalman/pdffiles/ladderpaper.pdf

The figures are missing though. You can probably find one with figures as well. Try googling it.

One last thing: make sure that your solution: 8.68 can fit in the corridor, e.i., what is its width? Is it smaller than 1 ?

Thanks for the link 8foxes and newzad. I now see where I went wrong. I just had the wrong equation. I shouldn't have assumed it just made a quarter circle as the line or ladder goes around the corner. According to the article 8foxes linked, it's x^2/3 + y^2/3 = L^2/3.

Now that I know the equation for sure, I can solve this answer.

Let me contribute to the discussion here. This is NOT a one-variable optimization problem. There are TWO parameters. In other words, there are TWO decision variables, L: Length and a: Angle with the horizontal. You can also go with Width and Angle, or even Length and Width. But my preference is L: Length and a: Angle. Here is my work:

W (Width) = 8sin(a) + cos(a) – sin(a)*cos(a)*L Area (L, W) = L*W Area (L, a) = [8sin(a) + cos(a)]*L – sin(a)*cos(a)*L^2

So, we have a non-linear optimization problem:

Max{ Area (L, a) = [8sin(a) + cos(a)]*L – sin(a)*cos(a)*L^2 Subject to: Width = 8sin(a) + cos(a) – sin(a)*cos(a)*L < 1 }

You can add additional constraints to narrow the feasible region such as: 8 < L < 5sqrt(5) <-- (See Fox 93)

One can probably use a math software to solve the above system. I believe there is a solution a little bigger than 8. Let’s see if we will be able to find it.

I think the problem is how can we know the rectangle pass or not pass.How can we be sure that a rectangle, width 1 and an area=5V5-5/2 could pass or not?

Newzad, Your (you and Ajit's) solution actually satisfies the above system. In other words, it is a feasible solution. Here is more details: Width (W) = 1.000 Length (L) = 8.68 Area (A) = 8.68 Angle (a) = 26.565 degrees Also note sin(2a) = 0.8 exactly! Very interesting observation: It is a 3-4-5 triangle!

So, 8.68 is A legitimate solution. I don't see any problem why it is not physically possible. But we can not be sure if there is a bigger area. (I don't think there is.) But one needs to prove that, 8.68 is optimum.

The answer should be smaller than 8.68 because if width=1, it can not pass. How can i say... think about this rectangle entering from the small coridor. ok i will draw, my english is not enougt to explain:) here it is http://i46.tinypic.com/2j2i1kp.gif

newzad is right, if you make the width 1, you have to have the length at 8 or below. Otherwise, you can't squeeze it in there just like his drawing shows.

If W=1 and L=8 the rect. cannot turn through the corner (which I think the problem asks for) but can only move “vertically”. To turn, Lmax ≤ √(8^2 – 1^2) = √63 = 3√7 = 7,937

If W=0 (straight line) then L = 1/sina + 8/cosa so to find Lmin → L' = 0 → a = arctan (1/8)^(1/3) = 26,565 deg → L= 11,18

I am sure that it is smaller than 8.68 :)

ReplyDeleteThe longest ladder if it were a straight line would be 5√5 and the angle it would make with the opposite wall would be 26.565 deg. With this geometry if we assume that the maximum width that will pass around the corner will be just under 1 then we could work out the maximum length of the ladder as approxly. 8.68034 and thus maximum area would be approxly. 8.68. Now someone please explain what's wrong with this argument?

ReplyDeleteAjit

I agree you Ajit, but i am not sure, So

ReplyDeleteif there is an answer it would be smaller or equal to 5V5-5/2=8.68

Here is the email I sent to fox1

ReplyDeleteOk... so here is what I did. If you just use a line first and no shape. You will find that the maximum length that can fit through that hole is 13. How do I know this? Well to understand this, first we start the line on the very left side standing up. then, as you move the bottom of the line to the right, the top of the line will naturally move down. This happens on the x axis and the y axis. And since they both move at the same rate, the line itself follows a circular path. It follows the bottom left side of a circle. The equation of a circle is (x-k)^2 + (y-h)^2 = r^2. Since the circle is going to have the center point (r,r), we can just change the formula to (x-r)^2 + (y-r)^2 = r^2. *NOTE* r is also the length of line *NOTE* Now, we can just plug in (8,1) to find out what equation of circle falls on that point. (8-r)^2 + (1-r)^2 = r^2. You basically end up with two answers here because there are two circles with center (r,r) that have that point lying on it's path. But I found that 13 was the answer because it was the one where the point (8,1) landed on it's bottom left part.

Now, that wasn't so bad. However the rectangle part is a little different and a little more difficult. If you just think about two lines running right next to each other instead, you can see that now you just have two lines running the same paths, except one is a little above the other. In fact, think of two circles, it's just a circle inside of another circle. One line follows the path of the big circle, and the other line follows the path of the circle inside of the bigger circle. Well we need to figure out two equations of these two circles. Since they BOTH lie on the same center (r,r) *NOTE* r is the radius of the big circle *NOTE*, we can say (x-r)^2 + (y-r)^2 = r^2 for the big circle like before, and (x-r)^2 + (y-r)^2 = L^2 (L being radius of smaller circle). Now we set x and y to (8,1) and we finally have our equation! This equation is based on the bigger circle that MUST lie on the x and y axis. Now you can just play with numbers to see what you get. (r-L) is our width of the rectangle and L is the length of the rectangle. r(r-L) would be your area that you get. And, you can use the equation above (L^2 = ....) and solve for L, the plug it into r(r-L) to get ONLY r's in that function. To maximize it, you just need to take the derivative of that with respect to r and set it equal to 0. But this is rather difficult and I'm not sure how I would do it.

But on to your question about if it could "really" fit in there. I don't see why you couldn't put it in there. Unless there is some kind of physics that disallows of it?

Thanks Michael

ReplyDeleteI think you mean this

http://en.wikipedia.org/wiki/Moving_sofa_problem

I hope this will help you

ReplyDeletehttp://www.jstor.org/stable/4145272?seq=1

or this;

ReplyDeletehttp://www.jstor.org/stable/1559028?seq=1

How does one go to the second page of this article by Moretti?

ReplyDeleteIt is a copyrighted article. You may need to purchase it online. Or your better chance is that you can borrow it from a university library. Here is another link:

ReplyDeletehttp://www1.american.edu/academic.depts/cas/mathstat/People/kalman/pdffiles/ladderpaper.pdf

The figures are missing though. You can probably find one with figures as well. Try googling it.

One last thing: make sure that your solution: 8.68 can fit in the corridor, e.i., what is its width? Is it smaller than 1 ?

fox1

Thanks.

ReplyDeleteIf the width is just smaller than 1 then my feeling is that the board (length=5√5-5/2~8.68) would just turn around and pass through.

Thanks for the link 8foxes and newzad. I now see where I went wrong. I just had the wrong equation. I shouldn't have assumed it just made a quarter circle as the line or ladder goes around the corner. According to the article 8foxes linked, it's x^2/3 + y^2/3 = L^2/3.

ReplyDeleteNow that I know the equation for sure, I can solve this answer.

http://wpcontent.answers.com/wikipedia/commons/d/d6/Evolute_of_astroid.gif

ReplyDeletethat's the curve it makes on the top right. It's not a quarter circle. Sorry about that guys. I'll correct my answer in a bit.

Let me contribute to the discussion here. This is NOT a one-variable optimization problem. There are TWO parameters. In other words, there are TWO decision variables, L: Length and a: Angle with the horizontal. You can also go with Width and Angle, or even Length and Width. But my preference is L: Length and a: Angle. Here is my work:

ReplyDeleteW (Width) = 8sin(a) + cos(a) – sin(a)*cos(a)*L

Area (L, W) = L*W

Area (L, a) = [8sin(a) + cos(a)]*L – sin(a)*cos(a)*L^2

So, we have a non-linear optimization problem:

Max{ Area (L, a) = [8sin(a) + cos(a)]*L – sin(a)*cos(a)*L^2

Subject to: Width = 8sin(a) + cos(a) – sin(a)*cos(a)*L < 1 }

You can add additional constraints to narrow the feasible region such as:

8 < L < 5sqrt(5) <-- (See Fox 93)

One can probably use a math software to solve the above system. I believe there is a solution a little bigger than 8. Let’s see if we will be able to find it.

-Polar Fox

I think the problem is how can we know the rectangle pass or not pass.How can we be sure that a rectangle, width 1 and an area=5V5-5/2 could pass or not?

ReplyDeleteOne minor correction:

ReplyDeleteMax{ ...

Subject to: Width = 8sin(a) + cos(a) – sin(a)*cos(a)*L < 1 }

Should be:

Max{ ...

Subject to: Width = 8sin(a) + cos(a) – sin(a)*cos(a)*L <= 1 }

Newzad,

ReplyDeleteYour (you and Ajit's) solution actually satisfies the above system. In other words, it is a feasible solution. Here is more details:

Width (W) = 1.000

Length (L) = 8.68

Area (A) = 8.68

Angle (a) = 26.565 degrees

Also note sin(2a) = 0.8 exactly! Very interesting observation: It is a 3-4-5 triangle!

So, 8.68 is A legitimate solution. I don't see any problem why it is not physically possible. But we can not be sure if there is a bigger area. (I don't think there is.) But one needs to prove that, 8.68 is optimum.

Thank you for your contributions.

-Polar Fox

The answer should be smaller than 8.68

ReplyDeletebecause if width=1, it can not pass.

How can i say...

think about this rectangle entering from the small coridor.

ok i will draw, my english is not enougt to explain:)

here it is

http://i46.tinypic.com/2j2i1kp.gif

newzad is right, if you make the width 1, you have to have the length at 8 or below. Otherwise, you can't squeeze it in there just like his drawing shows.

ReplyDeleteI found 8.18262902

ReplyDeleteI will share my solution, now i am at school library:)

sorry, calculation mistake

ReplyDeleteMinimise area = y(8x+y-1) / x

ReplyDeleteto obtain maximum area of 8.

-Yu

Stapa says:

ReplyDeleteLet begin with the limits of L , W

For W, probably 1≤ W ≤ 0

If W=1 and L=8 the rect. cannot turn through the corner (which I think the problem asks for) but can only move “vertically”. To turn, Lmax ≤ √(8^2 – 1^2) = √63 = 3√7 = 7,937

If W=0 (straight line) then L = 1/sina + 8/cosa so to find Lmin → L' = 0 → a = arctan (1/8)^(1/3) = 26,565 deg → L= 11,18

So 1≤ W ≤ 0 and 7,93 ≤ L ≤ 11,18

correct up to this ?