Yes I did that. But It’s effectively what he did when he used the formula. This is for pre calculus maths. The kids must be being taught the formula before learning differentiation.
@@yodaami its the solution of x = - b/2a ± Δx, wher Δx=(±√(b²-4ac))/2a more generally known as the quadratic formula (-b±√(b²-4ac))/2a .. without the Δx term
The most efficient use of fencing to capture maximum area, in a four-sided enclosure, is always a square. When one side of the enclosure is provided, and you have to construct only three sides, the most efficient way is to create part of a square. The answer to this problem is given as half of a square. To answer these problems, the total length of the short sides always equals the length of the long side, or y=X1+X2, and total fencing = 2(X1+X2). It's also true that you can rotate the square that your half square is taken from. The long side stays the same length, anything reduced from one of the short sides is added to the other. The area remains the same. At the extreme, one side goes to length zero and the other increase to length Y. Now the river and the fencing form an isosceles triangle with fenced sides Y. The area is 1/2 (2000)(2000) is 2,000,000 m^2 and the length of fencing is 4000m. If you do this, you elimanate the need for one of the fence posts. That's even more efficient.
Fence posts for chain link fencing are generally needed every 10 ft, so you'll still need approximately the same number of fence posts no matter what shape you build. There is a little more expense for corner and gate posts, due to a need for a concrete footing and a larger post to withstand the extra load, but you'll still have a number of posts as a function of the length, whether it's a corner or a line post.
If the corners require extra components to brace the corner post against being pulled over (often a diagonal brace from the top of the corner post downward along the fence to the base of the next post), this saves the 'extra equipment' of a corner. Another method if you don't really care about the shape so much, is fence in a semicircle with the river along the diameter. Or a 'semi-polygon' with each vertex a fence post. If there are enough, we might not need the extra bracing at each vertex and thus maximize the enclosed area.
I like this idea of rotating the square. :) My first thought was a half circle (before I saw it was supposed to be a rectangle), and so the half square actually makes perfect sense to me. However, I do not really see, how you knew it must be a half square in the first place?
@@deakzoltan2714Just like you instantly knew the biggest area with a certain circumference is a circle the biggest area of a rectangle is a square. And you instinctively wanted to place the river as the diagonal of the circle it's exactly the same with the square.
Take the 4000m of fencing, and roll it into a ball. Then use the Banach-Tarski paradox to double the amount of fencing; repeat endlessly until you have infinite fencing. Then take a 1L Gabriel's horn filled with paint, and since the 1L of paint can clearly fill the 1L Gabriel's horn that has infinite surface area, use that 1L of paint to paint the infinite amount of fencing. I hope that this helps. ;-) !!
This is an elegant way to solve this problem using the concepts of Maxima and Minima. Thanks so much for bringing back the memory of college freshman year.
Proving the formula for the vertex: the vertex is where the derivative of a function is zero and quadratic functions are written in the form of ax^2+bx+c and if we differentiate with respect to X we get 2ax+b=0 we rearrange for X to get the X-coordinate by taking away b from both sides so 2ax=-b then we divide both sides by 2a to get X=-b/2a
If Tom wasn't so fussy, he could fence in a semicircle with the river along the diameter, for an area of 2,546,479 (more than 25% more area). But that's outside the problem's restriction.
Happy that Tom realized he wants a rectangular fence. Look at fences in the countryside and notice they are rectangular in shape. Why is that? Do build one rectangular fence and a circular one with similar rigidity and you will understand why you don't want a circular one. Then add the wasted land as you can never buy a circular plot of land.
No river has straight banks so Step one - build it on inside of a meander of the river ... if it is a full ox bow and comes back on itself the enclosed area could be huge. Lateral thinking before math ...😊🇦🇺
There is a simpler way to analyse this problem that touches the heart of algebra: the observation that problems that can be described by the same equations have mathematically identical solutions irrespective of contexts, and therefore, it makes sense to analyse equations separated from contexts. It is well known that among all rectangles with a fixed perimeter, the largest area has a square. It can be proven using the multiplication formula (x-a)(x+a) = x^2 - a^2. It follows that the maximum of z*(b-z) is at z = b/2. The problem above can be formulated by a similar set of equations, y=4000 -z, 2*Area = z*(4000 -z) by substitution z=2x. Hence, z=2000, and x=1000, without knowing anything about parabolas.
@@garys5175 It is not because the context is different. Nevertheless, a maximal value of the expression z*(b-z) is for z=b/2, so if you manage to transform your equation into this, you immediately get the answer, which you then have to transform into your variables, z=2x.
Another way to think about it is to consider the case where he does need to fence all sides, which would be a square of side length 1000m. If we start with that square and remove the river side, we now have 1000m of fence to add to other sides to make the area bigger. To increase X by 1m takes 2m of fence. To increase Y by 1m takes 1m of fence. These values are fixed, considering the rectangular constraint.* Therefore, we increase our area the most by only increasing Y, leading to the same 1000 x 2000 area as the algebraic solution.
Geerkens' Law: For maximum area, total vertical fencing = total horizontal fencing = one half total fencing. Thus maximum area is 2000m * 1000m = 2,000,000 m^2.
When doing calculations it is easy to make misstakes, so in the real world you need some kinf of extra checking and convince others that you are correct before starting the huge task of setting up 4000 m of fence. Explaining that the solution presented in the video is correct to people that normally sets up fence posts could be a challenge. So I am suprised that noone suggested the iterative stupid-simple approach. Set a value for X, calculate Y and X*Y. Set a new value for X....se what gave the biggest area and repeat. Everyone will understand the math, and can see what X value resulted in the largest area.
Simple. Make the fenced area an half circle of radius 4000/π = 1273.2m, for an area of 2 546 479.1m. The problem shows a river, and don't forbid non-straight fences. There is nowhere that state that the fences MUST be straight.
The task specifically stated the largest rectangular area. If that was not stated the largest area would be a half circle. To bring the problem slightly closer to reality I do like that they specified it should be rectangular. If you started to sell circular plots of land you would end up with a mess as there would be lots of land in between the circles. Also the practicality of putting in fence posts is important. You can set up quite a few posts in a straight line without the need of support. You often only need to add support for the corner posts, and with close to 90 degree angle the supports are easy to set towards the closest posts. With a circular setup you are in for a major problem unless you can have a super stiff fence pushing between posts. A normal fence is done with fence wire. A wire or rope is great for pulling force, but a disaster for pushing force.
Dislike the use of a formula. Surely it is better to use the symmetry and largest rectangle will be the x value mid way between the roots. Roots are 0 and 2000 so x =1000. No remembering formula necessary.
And in the real world, you can't put a side fence in where the maths says it should go, because it happens to be where a river bed is, which is best avoided. But, it is OK another 20m away. Welcome to the world of constraints.
Maybe I'm stupid but I can't see how he could enclose an area that large with 4000 metres of fencing. The largest possible rectangle will be a square, surely? Each side of that square will have a length of 4000m/3, so 1333.3'm, giving an overall area of 1,777,777.7'm. Please explain how he could enclose 2,000,000m using the avaiable fencing while sticking to the rectangular rule. I don't get it.
It's a formula he developed himself, based on the situation given. The area of the rectangle is x*y. The perimeter of the rectangle is 2*x + 2*y. Since one of the y's is the river, and we don't need a fence along it, we're only interested in the fence length of 2*x + y, which equals 4000 ft. So this generates the constraint equation: 2*x + y = 4000 And the equation we're trying to maximize, A = x*(4000 - 2*x) Since this is a parabola, maximizing it is as simple as finding the vertex, which is as simple as averaging the two x-intercepts. In a general sense, this would be an application of calculus, to solve for where the slope (dA/dx) equals zero.
???? First, you're forgetting a few zeros, which you do redeem, but your max as x=y only applies when there are four sides - i.e. two "x"s and two "y"s. We don't have that here, so the x=y doesn't apply. Note that the actual answer is equivalent to two squares adjacent to each other of 1000 metres a side.
X and Y are sides of a square. The total length is 2x+2y or 4x or 4y. Therefore the maximum total area possible, when building all sides of an enclosure with 4000m of fencing is when each sides is 1000m. This means the maximum area is 1,000,000 m^2. Using a natural boundary for one side, doubles potential area. To build all four sides of a 4000m^2 pen with fencing, in the most efficient way, would mean that each side should have length = (4000^1/2) m in length. That means the sides would be 63.246 m long and the total length of fencing 252.98 m.
@@davidnewell3232 your math is extremely wonky and wrong. If you have 4000 m of fencing, and you use that to make 3 sides of a square, then each side will use exactly 1/3 of those 4000 m, ie 4000/3 m. So in that arrangement, X=Y=4000/3. Therefore the area of the square is (4000/3)*(4000/3)=1,777,777.777..... We can round that up to 1,777,778 square meters. However, that is not the maximum area we can get. If we consider instead that the edge of the river bisects a larger square that is 2000 m on a side for what would be a full square using 8000 m of fencing, knowing that we have only 4000 m of fencing, what we're really solving is what is the largest half of a square we can get. As the video shows, if the long side is 2000 m and the two short sides are 1000 m, then the area we enclose as half a square is 2,000,000 square meters. 2,000,000 sq m > 1,777,778 sq m.
An easier way is to create a square with 1000 meter sides. Take the 1000m length from the riverside / 2 = 500 and add 500m to each end of the square bottom, slide the square verticals to each end of the new line making it a shape 2000m long x 1000m wide. 2000 x 1000= 2,000,000 sq m
That may be easier, but you haven't explained why it works. Maybe you could have gotten a bigger area dividing that riverside fence by 3 and adding to each of the 3 sides. You don't actually know until you've done the analysis above, showing that this is the correct solution. Then it's easy to see alternate ways to visualize it.
No. The question was, "What's the largest rectangular area possible?" The answer must be an area in square meters, which "2x=y" is not. And your question seems to be implying that the area wouldn't depend upon how much fencing your have available.
I can stand this sort of phrasing. "What's the largest rectangular area possible?" You mean, in the universe or what? If you mean "What's the largest rectangular area possible that Tom can enclose with his fence?" then write that! You want me to do the calculation properly, have at least the curtesy to phrase the problem properly. It's not only pettiness but it will also avoid confusion. Mind you, this is not addressed at Mr. H but at whoever phrased this. (Unless it was Mr. H, then it will be addressed to him! 😅)
Other option: take the derivative of the area function (-4x+4000) and set to zero. The slope is zero at the vertex and it still yields X=1000!
Yes, to me its easier than finding the vertex and substituting it in.
Yes I did that. But It’s effectively what he did when he used the formula. This is for pre calculus maths. The kids must be being taught the formula before learning differentiation.
@@yodaami its the solution of x = - b/2a ± Δx, wher Δx=(±√(b²-4ac))/2a
more generally known as the quadratic formula (-b±√(b²-4ac))/2a .. without the Δx term
It's an algebra problem hotshot
If you are ready to do calculus you probably shouldn't be watching this channel.
The biggest area, if any shape was allowed, would be a semicircular fence.
😊 but not for the 99%
Can you prove that? Seems like it might be part of an ellipse.
Would have been my answer too. Nowhere is stated that it has to be rectangular.
Erm it is stated very clearly that it has to be a rectangle. Watch again.
My screen says “largest rectangular area possible”
I'm completely distracted by this guy's ability to write backhanded!
The most efficient use of fencing to capture maximum area, in a four-sided enclosure, is always a square. When one side of the enclosure is provided, and you have to construct only three sides, the most efficient way is to create part of a square. The answer to this problem is given as half of a square. To answer these problems, the total length of the short sides always equals the length of the long side, or y=X1+X2, and total fencing = 2(X1+X2). It's also true that you can rotate the square that your half square is taken from. The long side stays the same length, anything reduced from one of the short sides is added to the other. The area remains the same. At the extreme, one side goes to length zero and the other increase to length Y. Now the river and the fencing form an isosceles triangle with fenced sides Y. The area is 1/2 (2000)(2000) is 2,000,000 m^2 and the length of fencing is 4000m. If you do this, you elimanate the need for one of the fence posts. That's even more efficient.
Fence posts for chain link fencing are generally needed every 10 ft, so you'll still need approximately the same number of fence posts no matter what shape you build. There is a little more expense for corner and gate posts, due to a need for a concrete footing and a larger post to withstand the extra load, but you'll still have a number of posts as a function of the length, whether it's a corner or a line post.
If the corners require extra components to brace the corner post against being pulled over (often a diagonal brace from the top of the corner post downward along the fence to the base of the next post), this saves the 'extra equipment' of a corner. Another method if you don't really care about the shape so much, is fence in a semicircle with the river along the diameter. Or a 'semi-polygon' with each vertex a fence post. If there are enough, we might not need the extra bracing at each vertex and thus maximize the enclosed area.
I like this idea of rotating the square. :)
My first thought was a half circle (before I saw it was supposed to be a rectangle), and so the half square actually makes perfect sense to me. However, I do not really see, how you knew it must be a half square in the first place?
@@deakzoltan2714Just like you instantly knew the biggest area with a certain circumference is a circle the biggest area of a rectangle is a square. And you instinctively wanted to place the river as the diagonal of the circle it's exactly the same with the square.
That's exactly the real life example my Algebra teacher used except it was a barn instead of a river. 😄
that is a BIG barn
Take the 4000m of fencing, and roll it into a ball. Then use the Banach-Tarski paradox to double the amount of fencing; repeat endlessly until you have infinite fencing. Then take a 1L Gabriel's horn filled with paint, and since the 1L of paint can clearly fill the 1L Gabriel's horn that has infinite surface area, use that 1L of paint to paint the infinite amount of fencing. I hope that this helps. ;-) !!
Largest area enclosed with 4000 meters fencing material is 2,000,000 square meters.
eGreat problem, but I remember this problem from calculus where you take the first derivative of the area function. Thanks for sharing!!!
Love these videos. Excellently explained!
This is an elegant way to solve this problem using the concepts of Maxima and Minima. Thanks so much for bringing back the memory of college freshman year.
Proving the formula for the vertex: the vertex is where the derivative of a function is zero and quadratic functions are written in the form of ax^2+bx+c and if we differentiate with respect to X we get 2ax+b=0 we rearrange for X to get the X-coordinate by taking away b from both sides so 2ax=-b then we divide both sides by 2a to get X=-b/2a
If Tom wasn't so fussy, he could fence in a semicircle with the river along the diameter, for an area of 2,546,479 (more than 25% more area). But that's outside the problem's restriction.
That was exactly my first idea, but then I saw “rectangular area.” Oh well.
Happy that Tom realized he wants a rectangular fence. Look at fences in the countryside and notice they are rectangular in shape. Why is that? Do build one rectangular fence and a circular one with similar rigidity and you will understand why you don't want a circular one. Then add the wasted land as you can never buy a circular plot of land.
No river has straight banks so Step one - build it on inside of a meander of the river ... if it is a full ox bow and comes back on itself the enclosed area could be huge. Lateral thinking before math ...😊🇦🇺
There is a simpler way to analyse this problem that touches the heart of algebra: the observation that problems that can be described by the same equations have mathematically identical solutions irrespective of contexts, and therefore, it makes sense to analyse equations separated from contexts.
It is well known that among all rectangles with a fixed perimeter, the largest area has a square. It can be proven using the multiplication formula (x-a)(x+a) = x^2 - a^2. It follows that the maximum of z*(b-z) is at z = b/2. The problem above can be formulated by a similar set of equations, y=4000 -z, 2*Area = z*(4000 -z) by substitution z=2x. Hence, z=2000, and x=1000, without knowing anything about parabolas.
But the answer isn't a square
@@garys5175 It is not because the context is different. Nevertheless, a maximal value of the expression z*(b-z) is for z=b/2, so if you manage to transform your equation into this, you immediately get the answer, which you then have to transform into your variables, z=2x.
Another way to think about it is to consider the case where he does need to fence all sides, which would be a square of side length 1000m.
If we start with that square and remove the river side, we now have 1000m of fence to add to other sides to make the area bigger.
To increase X by 1m takes 2m of fence. To increase Y by 1m takes 1m of fence. These values are fixed, considering the rectangular constraint.*
Therefore, we increase our area the most by only increasing Y, leading to the same 1000 x 2000 area as the algebraic solution.
Geerkens' Law: For maximum area, total vertical fencing = total horizontal fencing = one half total fencing.
Thus maximum area is 2000m * 1000m = 2,000,000 m^2.
When doing calculations it is easy to make misstakes, so in the real world you need some kinf of extra checking and convince others that you are correct before starting the huge task of setting up 4000 m of fence.
Explaining that the solution presented in the video is correct to people that normally sets up fence posts could be a challenge.
So I am suprised that noone suggested the iterative stupid-simple approach.
Set a value for X, calculate Y and X*Y.
Set a new value for X....se what gave the biggest area and repeat.
Everyone will understand the math, and can see what X value resulted in the largest area.
I was always bad at this 😂 if a math problem deals with shapes in any way then it’s definitely harder for me to solve.
Thanks alot sir for this video you really helped me alot with this video
You are most welcome
So much easier with a bit of calculus.
Hi! You mentioned also “minimum possible area”, but with same fencing length. Wouldn’t minimum and maximum area be the same? Thanks.
Simple. Make the fenced area an half circle of radius 4000/π = 1273.2m, for an area of 2 546 479.1m.
The problem shows a river, and don't forbid non-straight fences. There is nowhere that state that the fences MUST be straight.
The task specifically stated the largest rectangular area. If that was not stated the largest area would be a half circle. To bring the problem slightly closer to reality I do like that they specified it should be rectangular. If you started to sell circular plots of land you would end up with a mess as there would be lots of land in between the circles.
Also the practicality of putting in fence posts is important. You can set up quite a few posts in a straight line without the need of support. You often only need to add support for the corner posts, and with close to 90 degree angle the supports are easy to set towards the closest posts. With a circular setup you are in for a major problem unless you can have a super stiff fence pushing between posts.
A normal fence is done with fence wire. A wire or rope is great for pulling force, but a disaster for pushing force.
Sir we may use AM-GM inequality
(2x + y)/2 >= √(2x * y)
So after silving the inequality
2000000 >= xy
Hence largest area is 2000000m²
Hi,
there are some three-digit numbers that are divisible by 8 and not divisible by 5؟
Please answer
Tanks
Is a semicircle more optimal for area quantity? It sure would be for utility.
Dislike the use of a formula. Surely it is better to use the symmetry and largest rectangle will be the x value mid way between the roots. Roots are 0 and 2000 so x =1000. No remembering formula necessary.
I liked this video before I saw it
ME: Oooohhh, yeah!!! Totally get it, thank you
Me after walking away: 2 million?? Did anyone else get that??
4 kilometers of straight rivet? Oh, it's a hypothetical river.
Make that 2 km?
@@wbrehaut yeah. my brain isn't as good as it used to be. I'm aging like fine milk.
Well if you only want to be a foot away from the bank😂
You can fence a lot more area if you set up the fence in a semicircle.
And in the real world, you can't put a side fence in where the maths says it should go, because it happens to be where a river bed is, which is best avoided. But, it is OK another 20m away. Welcome to the world of constraints.
How long would the sections have to be if you left a two meter space in front for a wooden gate?
Hello sir I want to talk with you
Maybe I'm stupid but I can't see how he could enclose an area that large with 4000 metres of fencing. The largest possible rectangle will be a square, surely? Each side of that square will have a length of 4000m/3, so 1333.3'm, giving an overall area of 1,777,777.7'm. Please explain how he could enclose 2,000,000m using the avaiable fencing while sticking to the rectangular rule. I don't get it.
It's rectangular shape but the problem states that there's no fence on the river side.
@@mrhtutoring Which is why I have divided the length of the fence into three sides, not four. You have not answered my question.
🤣👍📝
Sir can you tell where did this formula come from? (Formula to calculate the largest area possible).
It's a formula he developed himself, based on the situation given. The area of the rectangle is x*y. The perimeter of the rectangle is 2*x + 2*y. Since one of the y's is the river, and we don't need a fence along it, we're only interested in the fence length of 2*x + y, which equals 4000 ft.
So this generates the constraint equation: 2*x + y = 4000
And the equation we're trying to maximize, A = x*(4000 - 2*x)
Since this is a parabola, maximizing it is as simple as finding the vertex, which is as simple as averaging the two x-intercepts. In a general sense, this would be an application of calculus, to solve for where the slope (dA/dx) equals zero.
@@carultch thanks for the explanation
The largest area that is possible, is when x=y and we can certify that x•y=4000 m^2 4000^1/2=63,245
x=y=63,245
???? First, you're forgetting a few zeros, which you do redeem, but your max as x=y only applies when there are four sides - i.e. two "x"s and two "y"s. We don't have that here, so the x=y doesn't apply. Note that the actual answer is equivalent to two squares adjacent to each other of 1000 metres a side.
X and Y are sides of a square. The total length is 2x+2y or 4x or 4y. Therefore the maximum total area possible, when building all sides of an enclosure with 4000m of fencing is when each sides is 1000m. This means the maximum area is 1,000,000 m^2. Using a natural boundary for one side, doubles potential area. To build all four sides of a 4000m^2 pen with fencing, in the most efficient way, would mean that each side should have length = (4000^1/2) m in length. That means the sides would be 63.246 m long and the total length of fencing 252.98 m.
@@davidnewell3232 your math is extremely wonky and wrong.
If you have 4000 m of fencing, and you use that to make 3 sides of a square, then each side will use exactly 1/3 of those 4000 m, ie 4000/3 m.
So in that arrangement, X=Y=4000/3.
Therefore the area of the square is (4000/3)*(4000/3)=1,777,777.777.....
We can round that up to 1,777,778 square meters.
However, that is not the maximum area we can get. If we consider instead that the edge of the river bisects a larger square that is 2000 m on a side for what would be a full square using 8000 m of fencing, knowing that we have only 4000 m of fencing, what we're really solving is what is the largest half of a square we can get.
As the video shows, if the long side is 2000 m and the two short sides are 1000 m, then the area we enclose as half a square is 2,000,000 square meters.
2,000,000 sq m > 1,777,778 sq m.
1000x1000m
Did you miss that the river must be one side of the rectangle?
Aaaa!!
It's a square. 1000 metre x 1000 metre
No it’s not
An easier way is to create a square with 1000 meter sides. Take the 1000m length from the riverside / 2 = 500 and add 500m to each end of the square bottom, slide the square verticals to each end of the new line making it a shape 2000m long x 1000m wide. 2000 x 1000= 2,000,000 sq m
that's crazy
That may be easier, but you haven't explained why it works. Maybe you could have gotten a bigger area dividing that riverside fence by 3 and adding to each of the 3 sides. You don't actually know until you've done the analysis above, showing that this is the correct solution. Then it's easy to see alternate ways to visualize it.
@@BangkokBubonagliayour version yields a smaller area.
I would do a circle and fence of 2.546.479 square meters. That's 25% more!
"Using algebra in real life" means on the SAT/ACT. 😃
Um,,, a circle…
I don't get it. The 4000 is irrelevent. You are showing that 2*x = y is the answer for all such problems.
No. The question was, "What's the largest rectangular area possible?" The answer must be an area in square meters, which "2x=y" is not. And your question seems to be implying that the area wouldn't depend upon how much fencing your have available.
It annoys me that people use "meters square" and "square meters" interchangeably.
Here is the difference: A square plank with sides of 2 meters is a plank that's 2 meters square, but the area is 4 square meters.
They are interchangeable.
What about half-circle ?
"What's the largest rectangular area possible?" A circle?
@@wbrehaut I know. But it should be a "real example" and the goal was "max area". :) The best would be a bend of the river with radius 2000...
@@wbrehaut I know. But it should be a "real example" and the goal was "max area". :) The best would be a bend of the river with radius 2000...
I can stand this sort of phrasing. "What's the largest rectangular area possible?" You mean, in the universe or what?
If you mean "What's the largest rectangular area possible that Tom can enclose with his fence?" then write that!
You want me to do the calculation properly, have at least the curtesy to phrase the problem properly. It's not only pettiness but it will also avoid confusion.
Mind you, this is not addressed at Mr. H but at whoever phrased this. (Unless it was Mr. H, then it will be addressed to him! 😅)
Well, the question, along with the diagram is perfectly plain. I don't see how it could be any plainer without being pedantic.
@@user-hq4gu1jb3n It works on The Rio Grande so no worries :-)