Can we have a rectangle with perimeter = 1000 cm but area = 1 cm^2?

[@bprpmathbasics]

3:44 here’s the video How to complete the square kzbin.info/www/bejne/mZXPqmRubZVlhpo

[@bobh6728]

The answer is yes. The maximum is always a square, the minimum approaches 0. So since 1 is between the max and min, the answer is yes.

[@Neun_owo]

I love that when it comes to these types of questions, there are usually many ways to approach it. Since I learnt AM-GM recently fromt bprp, let's try to solve the question with it! We know that the rectangle's perimeter is 1000, It's length(x) and width(y) should add up to 500, so x+y=500. Using AM-GM inequality, (x+y)/2>=sqrt(xy), as x+y=500, 250>=sqrt(xy), xy

[@XenoFeisher1]

I'm pretty sure the sum and product of roots should not be used here cuz there's 2 variables which just so happen to be roots. Can you explain cuz I feel like I'm missing something here and you are right

[@Silvar55x]

When completing the square, I find it easier to add the term as a square, and keep it such on the left side to avoid going back-and-forth with expansion: ℓ² − 500ℓ = −1 ℓ² − 500ℓ + 250² = −1 + 250² (ℓ − 250)² = −1 + 62500 ℓ = 250 ± √62499

[@guidoreitsma8149]

It's very, very, very, very long and thin, just like Chile.

[@DemanaJaire]

Chile is so long and thin, people there only talk about south and north, but never about east and west. lol

[@IITDREAMresonance]

long rectangle?

[@MadaraUchihaSecondRikudo]

I usually like to go to the extremes to determine whether something is possible or not. The biggest possible area would be a square, that's easy to see. The smallest possible area would be at the limit where one side goes to 0, because the other side would go to 500 (a finite number) and 500 * 0 = 0. So since 1 is between 0 and 62500 (which is 250^2 which is the area of the square case), there must be some configuration of lengths to give you 1 as the area.

[@wes9627]

x+y=500; x=250+z; y=250-z; xy=(250+z)(250-z)=1; z^2=250^2-1; z=√(250^2-1); x=250+√(250^2-1) and y=250-√(250^2-1)

[@xinpingdonohoe3978]

One should be able to contain any chosen area within an arbitrary perimeter, I'd have thought. If we extend past rectangles, we could even contain such finite areas within infinite perimeters.

[@dhyeykulkarni3989]

wonderful idea! let's prove it?

[@peterhawes9680]

The perimeter has to be at least 4*(sqrt of area) with equality when it is a square.

[@charlesbarrow803]

Koch snowflake infinite perimeter finite volume

[@MrPaulc222]

Yes, but the long length would be very long and the short length, tiny, and it may have to be a very close approximation if in decimal. A more exact look: 2x + 2y = 1000 (equation 1) xy = 1 (equation 2) Rearrange it to 2x = 1,000 - 2y (2y(1000-2y) = 1 2000y - 4y^2 = 1 4y^2 - 2000y + 1 = 0 (2000+or-sqrt(4,000,000 - 4*4*2))/2 = y (2,000+or-sqrt(3,999,968)/2 = y (2,000+or-2*sqrt(999,992))/4 (2,000+or-4*sqrt(249,998) The square root approximates to 499.998, so the other direction would be approx 0.002 500+or-2*sqrt(6,249))/2 It would probably be possible to have an exact representation if leaving in surd form.

[@saqibmanzoor5106]

Length =499.998cm and width= 0.0002cm

[@mde189]

just set up two equations in system, you'll have: a+b=500, ab=1 Use Vietta's theorem (formulas) and set up the quadratic equation: x^2-500x+1=0 Find roots and you'll have the exact same answer without using calculus.

[@programaths]

Graphically, if the bottom left corner of the rectangle is on the origin and only the top right corner can move, then that corner is restricted to a line as such that 500=x+y or y=500-x. That line is continuous, so one of its point represent a rectangle of area=1cm.

[@AllanPoeLover]

細得就像一根針一樣

[@shadeblackwolf1508]

That parameter will have two long and two short sides. We also know the surface is 1 (unit does not matter). A x B is only 1 if B = 1/A. Therefore if a long side is x, then 2x + 2/x = 1000. Dividing by 2, x+1/x = 500. As X is the long side, that means 1/x is less than 1, so most of the contribution will be from X. Now, a calculator can give a precise number, a pc keyboard makes long comments easier, so i'm gonna guestimate that x is about 500 cm. That means to get to a surface of 1, the short side is about 0.002 cm, or 20 microns. That means that for all practical purposes, this thing would be flat. How good is this estimate? Well, we got a circumference of 1000.004 cm, meaning the deviation is 0.0004%. That is way more precise than the precision of the original question. Yes, i know you can get an exact solution by multiplying all terms by X and solving the quadratic, but phone keyboard

[@jamescollier3]

7:38 WHAT!!!? I've been searching Google shopping for the "tap to erase board".... unsubscribe .. 😅😊 kidding of course

[@nishisingh227]

Yeah you can. It would be a rectangle as thin as a bold line.

[@RylanceStreet]

This is the sort of problem where an approximation can give a sufficiently accurate answer far quicker than solving a quadratic equation. With such a large perimeter and small area the rectangle is obviously going to be very long and thin, and the length will be slightly less than half the perimeter divided by the area i.e. 500 cm. Now divide the area by the approximate length to get the width of 0.002 cm. Then subtract the width from the first estimated length to get 499.998 cm. Close enough for all practical purpose!

[@1nfius948]

i would've just drawn a straight line to represent that rectangle

[@Reader_Raccoon]

guys, I've got confused at 3:34 anyone knows a video where he explains it???

[@bprpmathbasics]

Here it is kzbin.info/www/bejne/mZXPqmRubZVlhpo

[@abc4953xyz]

He just wanted to simplify the left hand side in the (a - b)² format to make the equation easier to solve. l² - 2×½(500)×l + (½(500))² = (l - ½(500))² = (l - 250)²

[@Thagrynor]

Always love your videos, but the best part of this video was watching you try to draw that rectangle near the end lol .... I feel your pain with trying to draw things into even an approximation or example of what the scale sort of looks like when one dimension is ludicrously small relative to the other. Great video as always though. 🙂

[@sugarfrosted2005]

Yes non constructively. Maximized is a square, minimized is the degenerate case where height equals 0.

[@alclelalclel]

Let's call the sides of the rectangle a and b A = a*b, but the area is 1 so a*b=1 1/a = b a*(1/a)=1 for the perimeter we use the formula 2a + 2b = 1000 divide by 2 a + b = 500 a+1/a = 500 common denominator --> (a^2+1)/a = 500 --> a^2 + 1 = 500a a^2 - 500a + 1 = 500 we get two solutions a approx. 0.002 a= 499.998 they are eachothers reciprocals, so we can call one of them b a approx. 0.002 b= 499.98 and that's our solution

[@volcarona8401]

A = a×b = 1 (cm²) P = 2(a+b) = 1000 (cm) b = A/a → P = 2(a + 1/a) = 1000 2(a + 1/a) = 1000 | ÷2 a + 1/a = 500 | ×a a² + 1 = 500a | -500a a² - 500a + 1 = 0 Use -p/2 ± √(p²/4 - q): a = -(-500)/2 ± √((-500)²/4 - 1) a = 250 ± √(250000/4 -1) a = 250 ± √(62499) a_1 ≈ 499.998 a_2 ≈ 0.002 "Coincidentally", whichever value you pick for a, the other value is your b.

[@Metaverse-d9f]

Use Vieta's theorem

[@deosisx]

This is very easy after shape with infinite perimeter and finite area

[@itsphoenixingtime]

Lil john house type beat

[@dacianbonta2840]

obviously

[@a.tsuruya8]

Can we have a rectangle with perimeter = 1680cm, and area = 176331cm^2?

[@lagomoof]

Ah-ha. This feels more like an AndyMath kind of question. I don't think bprp leans into the memes quite so much!

[@-wx-78-]

(1680/4)±√[(1680/4)²−176331] = 420±√69 are the sides.

[@m.h.6470]

Solution: YES. 2a + 2b = 1000 a * b = 1 2a + 2b = 1000 |:2 a + b = 500 |-a b = 500 - a a * (500 - a) = 1 500a - a² = 1 |+a² -500a 0 = a² - 500a + 1 a = -(-500)/2 ± √((500/2)² - 1) a = 250 ± √(500²/4 - 4/4) a = 250 ± √(250000 - 4)/√4 a = 250 ± √249996/2 a ≅ 250 ± 499.996/2 a ≅ 250 ± 249.998 a₁ ≅ 0.002 a₂ ≅ 499.998 since a₁ + a₂ = 500 and b = 500 - a, if a takes the small value, b takes the large value and vice versa.

[@manudude02]

Thumbs down, rectangle not drawn to scale.

[@nimmira]

Anyone familiar with Photoshop, this is kinda like the "line marquee tool" (or "line selection tool") where you can choose a row or a column of pixels