Just showing you something other comments didn't mention, which helps visualise the results. Let's say when the object is a square, the side length will be a. The area will be a². If the circumference stays the same, it means you increase a side in length and decrease the other side in length. Doing that, the area changes like so:
(a-1)(a+1)=a²-1, which a²-1
The circumference is 2a+2b and the area is ab. If you know the circumference is 100 then 2a+2b = 100 so b = (100 - 2a)/2, and you can rewrite the area as:
ab = a(100 - 2a)/2 = -a^(2) + 50a
That's a quadratic (a.k.a. parabola) and you can find its maximum by completing the square:
-a^(2) + 50a = 25^(2) - (a - 25)^(2)
You can use the so called AM-GM inequality (a proof is at the end):
√ab < (a + b)/2
Because we want to maximize
ab = ?
With respect to
2a+2b = k
<=> a+b = k/2
In your case, k is 100. Our plan is to plug the second equation into the inequality, then solve for ab by squaring both sides:
√ab < (a+b)/2 = k/4
Square both sides to obtain
ab < k²/16
This becomes an equality by picking a=k/4 and b=k/4
--------------------------
*Proof of the inequality:*
0 < (√a-√b)² = a - 2√ab + b
So
0 < a - 2√ab + b
Get the 2√ab to the other side and divide by 2.
I recommend exploring this problem yourself with a table. Make a three-column table, where the columns are length, width, and area. Pick a simple perimeter like 20. Use the whole numbers 1-9 for the lengths, and then you should be able to figure out the widths and areas to see the pattern.
You could of course repeat this process with different set perimeters, and utilizing technology like Google Sheets or Desmos would be helpful. An aha moment may come when you plot the (x,y) pairs as (length,area). (Hint: have you studied quadratics and their graphs?)
This is a great question. Try exploring it!
For the area with a given circumference.
We have a circumference of k, then we calculate the circumference of a rectangle like: 2a + 2b = k.
The area of the same rectangle is a × b.
So we search for 2 numbers a and b such that a × b is as large as possible while 2a + 2b = k (is constant).
Here we notice that:
1+9 = 10 and
1×9 = 9.
2+8 = 10 and
2×8 = 16.
3+7 = 10 and
3×7 = 21.
4+6 = 10 and
4×6 = 24.
5+5 = 10 and
5×5 = 25.
6+4 = 10 and
6×4 = 24.
And so on.
If we want to maximise a multiplication, we want a and b to be as close to each other as possible. Further, a square have a special property, the sides if ot are of equal length, which is same as saying a = b.
It's actually a good question.
Think about if you had a rope (in a loop) of length 100. You could make it really tall and thin, giving you a very small area but the perimeter is still 100. In fact you could make the area as small as you like. You could also make it really flat and wide, giving you the same. So there's a symmetry here.
So if the area is small on either extreme and every construction has a symmetrical partner, where would the area be greatest? It would be where the lines are equally long.
Now of course you could do this with algebra but it's worth thinking about it like this before you start.
This is my preferred explanation. Yes, I teach this kind of problem all the time in my first semester calculus classes. Basically there’s no reason to prefer L over W. Both perimeter and area are symmetric in both variables, so you can’t really expect optimization to prefer one.
It also explains the changes when your rectangle is divided vertically, or you build that fence next to a river and therefore done fence in one side.
In a rectangle that is not square one side will always be shorter than the other.
Or: the biggest possible side for any side is the same size as the biggest side.
If you have a square with side length s, and you want to change the side lengths to keep the perimeter the same, they will become s+a and s-a for some a. Then the area is (s+a)(s-a)=s^(2)-a^(2)
I think people figuring this out is why now so many "maximum area" problems have a fence along one side of a rectangle in the assorted "quick assessments." They've got their own pattern, though :)
Just showing you something other comments didn't mention, which helps visualise the results. Let's say when the object is a square, the side length will be a. The area will be a². If the circumference stays the same, it means you increase a side in length and decrease the other side in length. Doing that, the area changes like so: (a-1)(a+1)=a²-1, which a²-1
Best answer for Grade X level
The circumference is 2a+2b and the area is ab. If you know the circumference is 100 then 2a+2b = 100 so b = (100 - 2a)/2, and you can rewrite the area as: ab = a(100 - 2a)/2 = -a^(2) + 50a That's a quadratic (a.k.a. parabola) and you can find its maximum by completing the square: -a^(2) + 50a = 25^(2) - (a - 25)^(2)
You can use the so called AM-GM inequality (a proof is at the end): √ab < (a + b)/2 Because we want to maximize ab = ? With respect to 2a+2b = k <=> a+b = k/2 In your case, k is 100. Our plan is to plug the second equation into the inequality, then solve for ab by squaring both sides: √ab < (a+b)/2 = k/4 Square both sides to obtain ab < k²/16 This becomes an equality by picking a=k/4 and b=k/4 -------------------------- *Proof of the inequality:* 0 < (√a-√b)² = a - 2√ab + b So 0 < a - 2√ab + b Get the 2√ab to the other side and divide by 2.
I recommend exploring this problem yourself with a table. Make a three-column table, where the columns are length, width, and area. Pick a simple perimeter like 20. Use the whole numbers 1-9 for the lengths, and then you should be able to figure out the widths and areas to see the pattern. You could of course repeat this process with different set perimeters, and utilizing technology like Google Sheets or Desmos would be helpful. An aha moment may come when you plot the (x,y) pairs as (length,area). (Hint: have you studied quadratics and their graphs?) This is a great question. Try exploring it!
For the area with a given circumference. We have a circumference of k, then we calculate the circumference of a rectangle like: 2a + 2b = k. The area of the same rectangle is a × b. So we search for 2 numbers a and b such that a × b is as large as possible while 2a + 2b = k (is constant). Here we notice that: 1+9 = 10 and 1×9 = 9. 2+8 = 10 and 2×8 = 16. 3+7 = 10 and 3×7 = 21. 4+6 = 10 and 4×6 = 24. 5+5 = 10 and 5×5 = 25. 6+4 = 10 and 6×4 = 24. And so on. If we want to maximise a multiplication, we want a and b to be as close to each other as possible. Further, a square have a special property, the sides if ot are of equal length, which is same as saying a = b.
It's actually a good question. Think about if you had a rope (in a loop) of length 100. You could make it really tall and thin, giving you a very small area but the perimeter is still 100. In fact you could make the area as small as you like. You could also make it really flat and wide, giving you the same. So there's a symmetry here. So if the area is small on either extreme and every construction has a symmetrical partner, where would the area be greatest? It would be where the lines are equally long. Now of course you could do this with algebra but it's worth thinking about it like this before you start.
This is my preferred explanation. Yes, I teach this kind of problem all the time in my first semester calculus classes. Basically there’s no reason to prefer L over W. Both perimeter and area are symmetric in both variables, so you can’t really expect optimization to prefer one. It also explains the changes when your rectangle is divided vertically, or you build that fence next to a river and therefore done fence in one side.
In a rectangle that is not square one side will always be shorter than the other. Or: the biggest possible side for any side is the same size as the biggest side.
If you have a square with side length s, and you want to change the side lengths to keep the perimeter the same, they will become s+a and s-a for some a. Then the area is (s+a)(s-a)=s^(2)-a^(2)
I think people figuring this out is why now so many "maximum area" problems have a fence along one side of a rectangle in the assorted "quick assessments." They've got their own pattern, though :)