Using Algebra in Real Life

[floridaflutery]

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!

[davidnewell]

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.

[jacqueshollands]

Love these videos. Excellently explained!

[haroldharris]

The biggest area, if any shape was allowed, would be a semicircular fence.

[KhanAcademyPoPolsku]

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.

[jungtran]

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.

[RichardMathsandStatistics]

Largest area enclosed with 4000 meters fencing material is 2, 000, 000 square meters.

[yaseenelhosseiny]

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

[MrMartinae]

eGreat problem, but I remember this problem from calculus where you take the first derivative of the area function. Thanks for sharing!!!

[MeMadeIt]

That's exactly the real life example my Algebra teacher used except it was a barn instead of a river. 😄

[JxH]

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. ;-) !!

[Stephn]

I'm completely distracted by this guy's ability to write backhanded!

[JqlGirl]

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.

[dilligafwoftam]

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 ...😊🇦🇺

[pietergeerkens]

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.

[moebro]

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.

[mikefochtman]

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.

[bpark]

You can fence a lot more area if you set up the fence in a semicircle.

[eagle]

Is a semicircle more optimal for area quantity? It sure would be for utility.

[Rohan]

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.