The last question is not well-defined. The answer to that is "you can make the area arbitrarily small". Just put two points arbitrary close to one vertex and the third point to the opposite side. Resulting triangle will be arbitrarily close to a line, so its area will be as small as you want. It's all "legit" too - because you avoid the "cheating" by turning the triangle into a line exactly, but it goes to show that there is no answer if we constrain ourselves only to a "proper" triangle.
@@loadstone5149So you chose your points on lines 3 and 4 to be arbitrarily close to the 3,4 vertex of the original triangle. Then you chose your points on the 5 line to be arbitrarily close to the 3,5 vertex. Now you have a triangle that is arbitrarily close to matching the 3 line.
The minimum triangle (question at the end) is the one, where two of the points are almost at the tip of the smaller angle and the the third point almost at the right angle. The area is miniscule, basically zero, as the triangle is almost a line by that point.
Similar solution: the point on the hypotenuse is close to 1 corner, the point on the vertical is close to the bottom, and the point on the bottom is anywhere. Since you can get as close as you want to the corners, the area approaches 0.
Another way of doing it is by finding the roots of the parabola. Doing so, you get that the area is smallest at x=0 or x=4. Assuming you want a non-zero answer, then there isn't an exact answer, because for any non-zero value you plug in to the area equation (x-4)*y, there will always be a smaller number that you could plug in.
@@qqqalo this has nothing to do with the parabola, we're talking about the puzzle he gave at the end, which is completely different to whatever problem you're trying to solve
Double the triangle, make a rectangle. Construct a perpendicular intersection within the interior of the rectangle perpendicular to the sides of the rectangle. There will be exactly one complete rectangle where the hypotenuse does not intersect the rectangle, until the interior corner of the rectangle hits the midpoint of the hypotenuse, at which point the single rectangle duplicates. This maximizes size of the rectangle, which is 1/4 the area of the larger rectangle defined by the shorter sides of the right triangle.
Man this was amazing, I solved this using differential calculus but this approach was superlative, and outstandingly genius, gave me a dopamine rush fr.
most of the optimization tasks in geometry end up in nice values like 1/2, 1/3 or 1/4 anyways i think there was some theorem that states if a+b=c, then a*b reaches its peak value when a=b, you could also see when a,b = c/2, apparently it's because of the AM-GM inequality
I had a smaller 3:4:5 triangle above the square with vertical h. This gave me the area if the square A = 4h - h^2(4/3). Seeing this was a (inverted) parabola, I differentiated the RHS and said at a max/min 4 - (8/3)h = 0. So h = 1.5 and is halfway the length of the vertical. The horizontal is (4/3) h = 6/3 =2 and the area is 1.5 x 2 =3 So apart from labelling everything differently I got the same sides and the same area for the rectangle. I’ll have to look for your calculus solution!
About the last question, can’t you just put two of your points on the 90 degree angle, and the 3rd point on the 5 long line? That would be an area of 0 because it’s a line, and if you want an actual triangle, move one of the points ever so little, and get an area infinitely close to 0. Right?
@@Crazy-mx9dc The problem is that you can get the area as close to zero as you like, by putting 2 of the points as close to a corner as you like. Any value you can name that's greater than zero, I can tweak it to be closer to zero. So, either zero *is* the minimum, or there is no minimum.
For largest area of triangle, dA/dx = 0, or 6x/4=3 or x=2. This way I don't have to remember equation of parabola or its peak point. (To make sure we get the highest area, not the lowest, one can take second derivative of A, if needed).
given the formula for area, A and that the maxinun is where dA/dX = 0. So differentiate the area formula and set to 0. A = 3x -(3/4)x^2 d/dx = 3 -(6/4)x set = 0 0 = 3-(6/4)x or -3(-4/6)=x 4/2 = x so x = 2
Question at end: given the constraints, I would put a vertex on the short leg just at the right angle, another on the hypotenuse just at the most acute angle, and the third vertex literally anywhere on the long leg. This forms a degenerate triangle with area 0. If we nudge the first two points ever so slightly up, the triangle is a very flat one with barely any area, and by moving these points closer to the vertices I mentioned, we can make this triangle;s area arbitrarily small, all the way to 0.
Same answer, different setup. I imagined the vertex on the hypotenuse as a bead on a string that I can slide left or right. That reduces the problem to one variable, since the bead's position determines all other vertices and lengths. I set the length of string left of the bead to "x", which makes the length of string right of the bead "5-x". Since the inner triangles are also 3-4-5 ratio, the rectangle is (3/5)(x) tall, and (4/5)(5-x) wide. Multiplied to get area, and found maximum from there.
I made x the ratio of the left piece (where you put x) and the full length (4); therefore the size of the rectangle is (3x) * (4 - 4x), which simplifies to 12 * (x - x^2). Setting the derivative to 0 gets: 0 = 12 * (1 - 2x) 0 = 1 - 2x x = 0.5 And the area is: 12 * (x - x^2) = 12 * (0.5 - 0.25) = 12 * 0.25 = 3
To me, the answer to the challenge question at the end is infinitesimally small, so I would say the limit of the area is zero. The points would approach 2 of the three verticies but never actually be at them.
Here’s an alternative way to express the height of the rectangle in terms of x: The small triangle in the bottom left is similar to the larger 3:4:5 triangle. Using the same labels as in the video, we have y/x=3/4, which gives y=3x/4. However, it might feel more intuitive (at least to me) to label the base of the small triangle as 4x instead of x. This way, it clearly suggests that the height of the small triangle (and therefore the rectangle) is 3x.
Alternative method (high school math competition 101): Let's say the rectangle has height a and width b, the triangle's hippo side has a slope of k, then we have 4 * k = 3 and a + b * k = 3 using arithmetic inequality: a * kb
Enjoyable problem. Not too hard but still requires some thought. I solved it exactly as our host, except I had forgotten the equation for the max/min of a parabola; probably because I learned it 40 years ago and hadn't used it since. However I had not forgotten how to take a derivative of a simple quadratic.
I wasn't sure (before I watched the video) that the rectangle was to share the right angle with the triangle, so after solving for that arrangement (area of the rectangle is maximally 1/2 the area of the triangle), I decided to make sure there wasn't a rectangle with one line on the hypotenuse which would be larger, but after some careful setup for an arbitrary non-obtuse triangle (vertices at (-a,0), (0,b), and (c,0)), it (of course) turns out that the maximal such rectangle is also exactly half the area of the triangle, with the upper vertices at the center points of edges ab and bc.
if you put 2 points on the right angle (equal to each other, but "touching" different sides) and the other point on the 5 side anywhere, wouldn't the area be 0? or would this question be better answered by figuring out the shortest distance from each corner to the opposite sides?
My intuition says that the biggest possible rectangle for most right triangles will have one side aligned along the hypothenuse. So, yes you spotted the question for rectangles that snuggle into the right angle, so you did what you planned to do However I am not convinced that this is actually the very biggest rectangle that could fit inside that triangle.
Imagine folding left triangle to the right and top one down. These triangles cover entire rectangle, and one of them sticks out - unless the triangles are exactly halves of the rectangle, thus minimizing their total area (and maximizing area of the rectangle therefore).
let the base of the triangle be "b" and the height of the triangle be "h" if we put the x and y axes like bprp did, and use the slope formula, we would get the line equation as y=hx/b now lets take a look at the rectangle's area, the area of the rectangle (A) = (b-x)y = (b-x)(hx/b) distributing hx/b to the two terms would give us A = -hx^2/b + hx if we use the same formula of bprp then we would get the x value at the maximum area i.e., x=-b/2a = -h/2(-h/b) = (-h/2)(b/-h) cancelling the "-h"s we would get x=b/2 now, to find y we must go back to the line equation y=hx/b substituting x=b/2, the equation becomes y=(h/b)(b/2) cancelling the "b"s we would get y=h/2 now, we should go back to the area of the rectangle and substitute x and y in there i.e., A = (b-x)y = (b-b/2)(h/2) simplifying, we get A=(b/2)(h/2) A=bh/4 we can write that as A=(1/2)(bh/2) A=(1/2)(Area of triangle) Therefore, Area of rectangle = half of Area of triangle So, answering your question, yes, the maximum of the rectangles area is always half of the original triangle :D
@@IMayBeAnAlien We can get to "yes" much easier than that. If you scale the triangle along the X axis, the area of every enclosed rectangle is scaled by the same factor, so the largest rectangle remains the one that's half the width and height of the triangle. Then you can scale along the Y axis, it's it's still true.
Watched the first minute, solved it by myself, skipped to the end, got a completely wrong answer of 3/8(9-sqrt(17)) or about 1.8288. I found that the final area would be (4-x)(3-y) and y=3x/4, then substituted 3x/4 for y in the first equation. I decided to graph y=(4-x)(3-3x/4) (for whatever reason) and y=3x/4 and see where they intersect to find the answer. I should've realized I was on the wrong path when they intersected at 2 points, one being completely outside of the triangle.
How do you know the biggest rectangle has the edges along x- and y-axes? We might find a bigger rectangle running along the hypotenuse, extending towards the right angle of the triangle.
@@yawninglion Analytic geometry told me that. 😉 Another way to approach the problem was shown in my other comment: three triangles beside that rectangle could be moved in such way so they cover it entirely and (if rectangle vertices aren't midpoints) there will be some extra triangular area outside the rectangle or overlap inside it. Consequently, these triangles have minimum total area when there's no overlap or extra; thus rectangle's maximum area is half of the big triangle's.
@@yawninglion Any point on the leg uniquely defines two rectangles: one with vertex at the right angle, another with side on hypotenuse. Those rectangles obviously have equal area (because they can be transformed into each other by shearing, which preserves area).
But I knew from early maths classes that the max rectangle (in a right-angle triangle) was always (x/2).(y/2) which here gives 3. Would I have lost points by declaring that?
Solution: When we put the triangle into a coordinate system, with the right angle and the zero point, the hypotenuse is either y = -3/4 * x + 3 or y = -4/3 * x + 4. This way, the target triangle is x * y or x * f(x), which is x * (-3/4 * x + 3) or x * (-4/3 * x + 4). Let's continue with the first: g(x) = x * (-3/4 * x + 3) g(x) = -3/4 * x² + 3x We are looking for the maximum value, so we need g'(x) = 0 and g''(x) < 0 g'(x) = -3/2 * x + 3 g''(x) = -3/2 → constant, therefore always a maximum g'(x) = 0: 0 = -3/2 * x + 3 |+3/2 * x 3/2 * x = 3 |*2/3 x = 2 So the maximum of g(x) = x * (-3/4 * x + 3) is at x = 2 The corresponding maximum area value is: g(2) = 2 * (-3/4 * 2 + 3) g(2) = -3/4 * 4 + 3 * 2 g(2) = -3 + 6 g(2) = 3 Therefore the maximum rectangle is 3 areal units
Coud you explain the unit of área you have used ?? " unit squared" as you said ?? What unit system is that ?? I was interested to apply this, in a chart "Power-rpm" (axis 'y' and 'x', KW vertical axis, rpm horizontal axis), but the unit of area non-standardized you used is disconcerning !!! It's imposible to square a KW unit.
Well, it's the same unit the original triangle had, but squared. if the triangle was 3cm - 4cm - 5cm, then, the area of the rectangle is 3cm². In this case, there wasn't any unit specified, 3 units - 4 units - 5 units right triangle. So unit squared for the answer.
@msolec2000 And why don't he put cm or cm² ?? from the begining. "Cm" is and standard unit in the international metric system. You can't go to the market to buy 3 units squared of something. And in my case, in my chart x-y Kw-rpm, I can't square any unit, are differents applications !!
@@marioalb9726 Yeah, but this is geometry. When you want to adapt that to your situation, you use the units you need. In other words, the important thing here is the calculation, not the unit. :)
Very tedious method ....compkete the square of 3×4 of triangle arms. 1/4 of square area is max in the triangle hence mwx area is 3..draw the picture to clear it... Now minimum triangle area in 1/2 of this quardent area that is 3/2
Sorry... you answered the wrong question. You made the extra assumption that one of the corners of the rectangle is in the right angle of the triangle.... and that is just an assumption. If you lay one of the sides of the rectangle on the hypotenuse of the triangle... you get the same size. I have not worked out yet what the maximum size is if the rectangle is ANYWHERE inside the triangle.... WIP.
Tried solving it before the video and got 3 as the biggest possible area. Let’s see if that’s right. Edit: I was right 😁 I used a different method, but I got the same solution.
@@torlumnitor8230 The area of the triangle is 1/2 * base * height. Substituting the values for this triangle results in: 1/2 * 3 * 4, which is also 3 * 2, or 6. The area of the square is 3, so the area of the triangle is twice that area.
Real life application: under the eaves of a roofline, I want to install the largest possible air duct: what’s the largest duct I could possibly use? I’ve done this exact exercise numerous times graphically (because math hurts my head) using drafting software. But real world: the answer is your outer diameter and be sure to reduce your results by .25-.5” or you’ll never get it installed! (Ask me how I know!) 🫣
The last question is not well-defined. The answer to that is "you can make the area arbitrarily small". Just put two points arbitrary close to one vertex and the third point to the opposite side. Resulting triangle will be arbitrarily close to a line, so its area will be as small as you want. It's all "legit" too - because you avoid the "cheating" by turning the triangle into a line exactly, but it goes to show that there is no answer if we constrain ourselves only to a "proper" triangle.
Each vertex must be on a different side
@@loadstone5149So you chose your points on lines 3 and 4 to be arbitrarily close to the 3,4 vertex of the original triangle. Then you chose your points on the 5 line to be arbitrarily close to the 3,5 vertex. Now you have a triangle that is arbitrarily close to matching the 3 line.
This is the answer I came up with.
The minimum triangle (question at the end) is the one, where two of the points are almost at the tip of the smaller angle and the the third point almost at the right angle. The area is miniscule, basically zero, as the triangle is almost a line by that point.
Similar solution: the point on the hypotenuse is close to 1 corner, the point on the vertical is close to the bottom, and the point on the bottom is anywhere. Since you can get as close as you want to the corners, the area approaches 0.
Another way of doing it is by finding the roots of the parabola. Doing so, you get that the area is smallest at x=0 or x=4.
Assuming you want a non-zero answer, then there isn't an exact answer, because for any non-zero value you plug in to the area equation (x-4)*y, there will always be a smaller number that you could plug in.
One could say infinitesimally small.
@@qqqalo this has nothing to do with the parabola, we're talking about the puzzle he gave at the end, which is completely different to whatever problem you're trying to solve
Double the triangle, make a rectangle. Construct a perpendicular intersection within the interior of the rectangle perpendicular to the sides of the rectangle. There will be exactly one complete rectangle where the hypotenuse does not intersect the rectangle, until the interior corner of the rectangle hits the midpoint of the hypotenuse, at which point the single rectangle duplicates. This maximizes size of the rectangle, which is 1/4 the area of the larger rectangle defined by the shorter sides of the right triangle.
Man this was amazing, I solved this using differential calculus but this approach was superlative, and outstandingly genius, gave me a dopamine rush fr.
most of the optimization tasks in geometry end up in nice values like 1/2, 1/3 or 1/4 anyways
i think there was some theorem that states if a+b=c, then a*b reaches its peak value when a=b, you could also see when a,b = c/2, apparently it's because of the AM-GM inequality
I had a smaller 3:4:5 triangle above the square with vertical h. This gave me the area if the square A = 4h - h^2(4/3). Seeing this was a (inverted) parabola, I differentiated the RHS and said at a max/min 4 - (8/3)h = 0. So h = 1.5 and is halfway the length of the vertical. The horizontal is (4/3) h = 6/3 =2 and the area is 1.5 x 2 =3 So apart from labelling everything differently I got the same sides and the same area for the rectangle. I’ll have to look for your calculus solution!
You could use max min. Or, you could just take a 45, 45, 90 triangle, with a square inscribed in it, and stretch it by 1/3.
About the last question, can’t you just put two of your points on the 90 degree angle, and the 3rd point on the 5 long line?
That would be an area of 0 because it’s a line, and if you want an actual triangle, move one of the points ever so little, and get an area infinitely close to 0. Right?
that might be considered cheating because the limit of the area is zero, and not the actual area? please correct me if i'm wrong 😅
@@zoinkerspoinker4326maybe if we get this to be top comment, we’ll get an answer
@@zoinkerspoinker4326 I’m just a 15 year old student, I’ve got no idea if that’s right or wrong 😅
@@Crazy-mx9dc The problem is that you can get the area as close to zero as you like, by putting 2 of the points as close to a corner as you like. Any value you can name that's greater than zero, I can tweak it to be closer to zero. So, either zero *is* the minimum, or there is no minimum.
@@jursamaj that’s how far I understood it, but still thanks
For largest area of triangle, dA/dx = 0, or 6x/4=3 or x=2.
This way I don't have to remember equation of parabola or its peak point.
(To make sure we get the highest area, not the lowest, one can take second derivative of A, if needed).
given the formula for area, A
and that the maxinun is where dA/dX = 0. So differentiate the area formula and set to 0.
A = 3x -(3/4)x^2
d/dx = 3 -(6/4)x
set = 0
0 = 3-(6/4)x
or
-3(-4/6)=x
4/2 = x
so
x = 2
Question at end: given the constraints, I would put a vertex on the short leg just at the right angle, another on the hypotenuse just at the most acute angle, and the third vertex literally anywhere on the long leg. This forms a degenerate triangle with area 0. If we nudge the first two points ever so slightly up, the triangle is a very flat one with barely any area, and by moving these points closer to the vertices I mentioned, we can make this triangle;s area arbitrarily small, all the way to 0.
Same answer, different setup.
I imagined the vertex on the hypotenuse as a bead on a string that I can slide left or right. That reduces the problem to one variable, since the bead's position determines all other vertices and lengths. I set the length of string left of the bead to "x", which makes the length of string right of the bead "5-x".
Since the inner triangles are also 3-4-5 ratio, the rectangle is (3/5)(x) tall, and (4/5)(5-x) wide. Multiplied to get area, and found maximum from there.
I made x the ratio of the left piece (where you put x) and the full length (4); therefore the size of the rectangle is (3x) * (4 - 4x), which simplifies to 12 * (x - x^2).
Setting the derivative to 0 gets:
0 = 12 * (1 - 2x)
0 = 1 - 2x
x = 0.5
And the area is:
12 * (x - x^2) =
12 * (0.5 - 0.25) =
12 * 0.25 =
3
3 sq units
equation for hypotenuse is: y = (3/4)xo and xo is: xo = 4.- x and 0
To me, the answer to the challenge question at the end is infinitesimally small, so I would say the limit of the area is zero. The points would approach 2 of the three verticies but never actually be at them.
Here’s an alternative way to express the height of the rectangle in terms of x:
The small triangle in the bottom left is similar to the larger 3:4:5 triangle. Using the same labels as in the video, we have y/x=3/4, which gives y=3x/4. However, it might feel more intuitive (at least to me) to label the base of the small triangle as 4x instead of x. This way, it clearly suggests that the height of the small triangle (and therefore the rectangle) is 3x.
Alternative method (high school math competition 101):
Let's say the rectangle has height a and width b, the triangle's hippo side has a slope of k, then we have 4 * k = 3 and a + b * k = 3
using arithmetic inequality: a * kb
Enjoyable problem. Not too hard but still requires some thought. I solved it exactly as our host, except I had forgotten the equation for the max/min of a parabola; probably because I learned it 40 years ago and hadn't used it since. However I had not forgotten how to take a derivative of a simple quadratic.
It’s obvious that x = 1/2l is the solution (otherwise there would be two solutions to the problem found by exchanging xl-x)
I wasn't sure (before I watched the video) that the rectangle was to share the right angle with the triangle, so after solving for that arrangement (area of the rectangle is maximally 1/2 the area of the triangle), I decided to make sure there wasn't a rectangle with one line on the hypotenuse which would be larger, but after some careful setup for an arbitrary non-obtuse triangle (vertices at (-a,0), (0,b), and (c,0)), it (of course) turns out that the maximal such rectangle is also exactly half the area of the triangle, with the upper vertices at the center points of edges ab and bc.
You have to define "large" first. Area? Perimiter? Height?
if you put 2 points on the right angle (equal to each other, but "touching" different sides) and the other point on the 5 side anywhere, wouldn't the area be 0? or would this question be better answered by figuring out the shortest distance from each corner to the opposite sides?
There is no minimum area of triangle. Because it can't be 0 but it can be 0,00......001.
I love the Christmas tree on the right of the whiteboard. Can you post that somewhere, please, or link to the source of it belongs to someone else
My intuition says that the biggest possible rectangle for most right triangles will have one side aligned along the hypothenuse.
So, yes you spotted the question for rectangles that snuggle into the right angle, so you did what you planned to do
However I am not convinced that this is actually the very biggest rectangle that could fit inside that triangle.
Light work
Max area is 3 and occurs when b = 2 and h = 3/2
Simplify sqrt 464
@ 2sqrt(116)
4sqrt(29)
By generalizing any right triangle, the area of the rectangle will always equal ab/4.
And the area of the inscribed circle is only slightly larger at pi...
1 1 √2 triangle is 1/4, scale it by 12, you get 3
Imagine folding left triangle to the right and top one down. These triangles cover entire rectangle, and one of them sticks out - unless the triangles are exactly halves of the rectangle, thus minimizing their total area (and maximizing area of the rectangle therefore).
holy shit its wx 78
What happens if "WE" decide to place the rectangle adjacent to the HYPOTENUSE? Is the answer still 3?
Could this be generalized, and will the general maximum area of an inscribed square always be half the area of the original triangle?
let the base of the triangle be "b" and the height of the triangle be "h"
if we put the x and y axes like bprp did, and use the slope formula, we would get the line equation as y=hx/b
now lets take a look at the rectangle's area,
the area of the rectangle (A) = (b-x)y = (b-x)(hx/b)
distributing hx/b to the two terms would give us
A = -hx^2/b + hx
if we use the same formula of bprp then we would get the x value at the maximum area
i.e., x=-b/2a = -h/2(-h/b) = (-h/2)(b/-h)
cancelling the "-h"s we would get
x=b/2
now, to find y we must go back to the line equation y=hx/b
substituting x=b/2, the equation becomes y=(h/b)(b/2)
cancelling the "b"s we would get
y=h/2
now, we should go back to the area of the rectangle and substitute x and y in there
i.e., A = (b-x)y = (b-b/2)(h/2)
simplifying, we get
A=(b/2)(h/2)
A=bh/4
we can write that as
A=(1/2)(bh/2)
A=(1/2)(Area of triangle)
Therefore, Area of rectangle = half of Area of triangle
So, answering your question, yes, the maximum of the rectangles area is always half of the original triangle :D
@@IMayBeAnAlien We can get to "yes" much easier than that. If you scale the triangle along the X axis, the area of every enclosed rectangle is scaled by the same factor, so the largest rectangle remains the one that's half the width and height of the triangle. Then you can scale along the Y axis, it's it's still true.
Yes the general case is true (the triangles outside the rectangle are similar, so the maximum must be when dimensions are equal)
❤ other method
taking vertices as (0,0) , (4,0) and (0,3)
(x/4)+(y/3) = 1
by AM -GM inequality
x/4+y/3 > = 2 sqrt(xy/12)
1 > = xy/3
Max area is 3
Can you explain to me where (x/4)+(y/3)=1 came from?
@assassin01620 intercept form of straight line in co - ordinate geometry
@raghvendrasingh1289 wow. How have I never heard of intercept form before? Thanks
@@assassin01620 👍
Watched the first minute, solved it by myself, skipped to the end, got a completely wrong answer of 3/8(9-sqrt(17)) or about 1.8288. I found that the final area would be (4-x)(3-y) and y=3x/4, then substituted 3x/4 for y in the first equation. I decided to graph y=(4-x)(3-3x/4) (for whatever reason) and y=3x/4 and see where they intersect to find the answer. I should've realized I was on the wrong path when they intersected at 2 points, one being completely outside of the triangle.
but what if you lay the rectange on the hypotenuse?
How do you know the biggest rectangle has the edges along x- and y-axes? We might find a bigger rectangle running along the hypotenuse, extending towards the right angle of the triangle.
Actually no, rectangle along hypotenuse have exactly the same maximum area (two vertices are midpoints of the triangle's legs again). 😉
@-wx-78- You'd still need to argue why that's not bigger than the one presented in the video.
@@yawninglion Analytic geometry told me that. 😉
Another way to approach the problem was shown in my other comment: three triangles beside that rectangle could be moved in such way so they cover it entirely and (if rectangle vertices aren't midpoints) there will be some extra triangular area outside the rectangle or overlap inside it. Consequently, these triangles have minimum total area when there's no overlap or extra; thus rectangle's maximum area is half of the big triangle's.
Well, I know the largest rectangle is indeed half of the triangle - but the argument in the video is incomplete
@@yawninglion Any point on the leg uniquely defines two rectangles: one with vertex at the right angle, another with side on hypotenuse. Those rectangles obviously have equal area (because they can be transformed into each other by shearing, which preserves area).
Based on what I got for my final answer of the question in the end. I don’t feel like typing it in, it’s way too long.
My answer is
1536/625 or 2.4576
But I knew from early maths classes that the max rectangle (in a right-angle triangle) was always (x/2).(y/2) which here gives 3. Would I have lost points by declaring that?
i think you would have had to show your working.
And that maximum area is half the area of the 3-4-5 triangle.
i had an intuition that it would be nine based on the geometry, is it correct to say that all of those triangles are 3,4,5 ?
I suspect you are right. I was guessing all the triangles were scaled versions of each other
Yes, all the triangles are similar. Their 3 sides are all parallel between triangles.
The "so-called" derivative!
Is the biggest rectangle inscribed in a triangle always going to have half the area of the triangle?
Solution:
When we put the triangle into a coordinate system, with the right angle and the zero point, the hypotenuse is either y = -3/4 * x + 3 or y = -4/3 * x + 4.
This way, the target triangle is x * y or x * f(x), which is x * (-3/4 * x + 3) or x * (-4/3 * x + 4).
Let's continue with the first:
g(x) = x * (-3/4 * x + 3)
g(x) = -3/4 * x² + 3x
We are looking for the maximum value, so we need g'(x) = 0 and g''(x) < 0
g'(x) = -3/2 * x + 3
g''(x) = -3/2 → constant, therefore always a maximum
g'(x) = 0:
0 = -3/2 * x + 3 |+3/2 * x
3/2 * x = 3 |*2/3
x = 2
So the maximum of g(x) = x * (-3/4 * x + 3) is at x = 2
The corresponding maximum area value is:
g(2) = 2 * (-3/4 * 2 + 3)
g(2) = -3/4 * 4 + 3 * 2
g(2) = -3 + 6
g(2) = 3
Therefore the maximum rectangle is 3 areal units
Why is the other one blue 😢😢😢😢
Last one: 0.
Three units square is 9 square units
Coud you explain the unit of área you have used ??
" unit squared" as you said ??
What unit system is that ??
I was interested to apply this, in a chart "Power-rpm" (axis 'y' and 'x', KW vertical axis, rpm horizontal axis), but the unit of area non-standardized you used is disconcerning !!! It's imposible to square a KW unit.
Well, it's the same unit the original triangle had, but squared. if the triangle was 3cm - 4cm - 5cm, then, the area of the rectangle is 3cm². In this case, there wasn't any unit specified, 3 units - 4 units - 5 units right triangle. So unit squared for the answer.
@msolec2000
And why don't he put cm or cm² ?? from the begining.
"Cm" is and standard unit in the international metric system.
You can't go to the market to buy 3 units squared of something.
And in my case, in my chart x-y Kw-rpm, I can't square any unit, are differents applications !!
@@marioalb9726 Yeah, but this is geometry. When you want to adapt that to your situation, you use the units you need. In other words, the important thing here is the calculation, not the unit. :)
Very tedious method ....compkete the square of 3×4 of triangle arms. 1/4 of square area is max in the triangle hence mwx area is 3..draw the picture to clear it...
Now minimum triangle area in 1/2 of this quardent area that is 3/2
smallest triangle is infinitely close to zero.
Sorry... you answered the wrong question.
You made the extra assumption that one of the corners of the rectangle is in the right angle of the triangle.... and that is just an assumption.
If you lay one of the sides of the rectangle on the hypotenuse of the triangle... you get the same size.
I have not worked out yet what the maximum size is if the rectangle is ANYWHERE inside the triangle.... WIP.
Tried solving it before the video and got 3 as the biggest possible area. Let’s see if that’s right.
Edit: I was right 😁 I used a different method, but I got the same solution.
Which is surprisingly being half of the big triangle area
How did you come to that conclusion?
@@torlumnitor8230I have contacts in high places
@@torlumnitor8230 The area of the triangle is 1/2 * base * height. Substituting the values for this triangle results in:
1/2 * 3 * 4, which is also 3 * 2, or 6.
The area of the square is 3, so the area of the triangle is twice that area.
He teaches us the max, in the shortest time😂!
Real life application: under the eaves of a roofline, I want to install the largest possible air duct: what’s the largest duct I could possibly use? I’ve done this exact exercise numerous times graphically (because math hurts my head) using drafting software. But real world: the answer is your outer diameter and be sure to reduce your results by .25-.5” or you’ll never get it installed! (Ask me how I know!) 🫣