I taught and wrote about mathematics for the best part of 40 years. But I never had anything close to your ability to make complex ideas both interesting and intelligible. Your videos are the most informative and interesting on the internet.
As an interesting extension which amounts to the same thing you might ask "Prove that the triangle with a fixed hypotenuse is isosceles." This can be proved in almost exactly the same way as you have done in this video. The idea of a relationship between perimeter and maximum area is very interesting. I am wondering I can show that for a fixed perimeter that the maximum area will be a circle. At least you got me thinking!
I have a very simole answer to this. Imagine a circle with a diameter of 5 cm. Pick a random point on the circle and connect that dot with the two end points of the diamater. Now you have a right triangle with a hypotenuse of 5 cm. Now it’s easy to see and calculate what’s the biggest possible right triangle is with a 5 cm hypotenuse and that is when the altitude is the radius of the circle. Which is 5/2=2.5 cm. Any higher and the triangle won’t be a right triangle anymore. So the answer is simply 1/2*5*2,5= 6,25 sq cm.
❤ elementary solution (without calculus) We have to maximize A = xy/2 Consider inequality (x - y)^2 > = 0 , equality holds iff x= y x^2+y^2 > = 2xy 25 > = 2xy xy < = 25/2 , equality holds iff x = y ( x = 5/√2 , y = 5/√2 ) Maximum area = 25/4 sq units
x^2+y^2=25 Pythagoras theorem, sqrt((x^2+y^2)/2)>=sqrt(xy) - proved inequality between averages Raise both sides to the power of 2 (x^2+y^2)/2>=xy devide both sides by two and replace x^2+y^2 with 25 25/4>=1/2xy , 1/2xy is our area, so it equals to 6.25.
I initially used the method from the video but realized in the middle of my solution that a cleaner trigonometric approach is possible since the Pythagorean equation represents a circle. We can express x as 5cos(v) and y as 5sin(v), where v is the angle at the bottom-right vertex. The triangle’s area is (5cos(v) × 5sin(v)) / 2, which simplifies to (25/4)sin(2v) using the identity 2cos(v)sin(v) = sin(2v). Since sin(2v) reaches a maximum value of 1 for 0 < v < 90, the maximum area is 25/4 cm².
....Good day to you Newton, Nice to see that you are enthusiastically tackling math topics, especially when it comes to optimization problems. I can well imagine that the same enthusiasm can make you confused for a moment, because you think you can think faster than you actually can as a human being; at least this is what I recognize in myself oh so well! Concerning this problem, I would have emphasized at the end (as a valuable conclusion) that the maximum area is always reached with an isosceles triangle, and is therefore half the area of the imaginary square with (equal) sides 5/sqrt(2) and fixed diagonal 5. For example, if you take a fixed perimeter for rectangles in different side lengths, you will reach the maximum area with a square with the same perimeter, and thus for isosceles triangles the same... Thank you Newton and stay well, Jan-W p.s. Also think of constraints and objective function...
Sometimes I think I could add and subtract as quickly as I could but it doesn't happen. You are correct, emphasizing the general idea of maximum area being half the imaginary square would have been a quick tip even from the beginning since the largest rectangle is always a square.
@@PrimeNewtons I already shared it with my friends. I suggest please upload more optimization problems because a lot of students are having a hard time with this particular lesson. Thanks a lot!!
This is a fun puzzle if you don't over-think it. I was able to solve in 30 sec by knowing the largest rectangle bounded by its diagonal is formed by equal side lengths. The side length of such a square is the diagonal over root2. Multiply the side lengths and take half to get the largest triangle.
Taking this a bit further, the area is 1/2 x*y, but x=y. So area = 1/2 (x^2). Also, since it's an isocelese triangle x^2 = 1/2 (hyp)^2. So we get area = 1/2*(1/2*(5cm)^2) = 25/4
It is! Have you proved it? I am very interested in the relationship between symmetry and area. As you say, the largest area you can get in a four-sided figure, for a fixed perimeter, or a fixed diagonal, it seems, is a square, which is the most symmetric quadrilateral. I wonder how far this can be extended.
I had a different approach. Given that for the same hypotenuse, 45° 45° 90° triangle will have the largest area. By applying the property of the special triangle, I can do (5/√2)^2/2, that will be the product of the length and width/2, which is equal to 6.25
Good point, as we're looking for the maximum value of an expression on a positive interval, meaning the squared expression will reach its maximum value at the same x-coordinate.
why did you equate the denominator of the first derivative to zero when you were trying to find the CNs. The values of x which make the denominator zero are the ones which make it undefined, correct? so why didnt you just equate the numerator
![How to Solve ANY Optimization Problem [Calc 1]](/img/n.gif)
I taught and wrote about mathematics for the best part of 40 years. But I never had anything close to your ability to make complex ideas both interesting and intelligible. Your videos are the most informative and interesting on the internet.
You are my favourite math tutor on youtube
As an interesting extension which amounts to the same thing you might ask "Prove that the triangle with a fixed hypotenuse is isosceles."
This can be proved in almost exactly the same way as you have done in this video. The idea of a relationship between perimeter and maximum area is very interesting. I am wondering I can show that for a fixed perimeter that the maximum area will be a circle. At least you got me thinking!
I have a very simole answer to this.
Imagine a circle with a diameter of 5 cm.
Pick a random point on the circle and connect that dot with the two end points of the diamater.
Now you have a right triangle with a hypotenuse of 5 cm.
Now it’s easy to see and calculate what’s the biggest possible right triangle is with a 5 cm hypotenuse and that is when the altitude is the radius of the circle. Which is 5/2=2.5 cm. Any higher and the triangle won’t be a right triangle anymore.
So the answer is simply 1/2*5*2,5= 6,25 sq cm.
❤
elementary solution (without calculus)
We have to maximize A = xy/2
Consider inequality
(x - y)^2 > = 0 , equality holds iff x= y
x^2+y^2 > = 2xy
25 > = 2xy
xy < = 25/2 , equality holds iff x = y ( x = 5/√2 , y = 5/√2 )
Maximum area = 25/4 sq units
x^2+y^2=25 Pythagoras theorem,
sqrt((x^2+y^2)/2)>=sqrt(xy) - proved inequality between averages
Raise both sides to the power of 2
(x^2+y^2)/2>=xy devide both sides by two and replace x^2+y^2 with 25
25/4>=1/2xy , 1/2xy is our area, so it equals to 6.25.
I initially used the method from the video but realized in the middle of my solution that a cleaner trigonometric approach is possible since the Pythagorean equation represents a circle. We can express x as 5cos(v) and y as 5sin(v), where v is the angle at the bottom-right vertex. The triangle’s area is (5cos(v) × 5sin(v)) / 2, which simplifies to (25/4)sin(2v) using the identity 2cos(v)sin(v) = sin(2v). Since sin(2v) reaches a maximum value of 1 for 0 < v < 90, the maximum area is 25/4 cm².
....Good day to you Newton, Nice to see that you are enthusiastically tackling math topics, especially when it comes to optimization problems. I can well imagine that the same enthusiasm can make you confused for a moment, because you think you can think faster than you actually can as a human being; at least this is what I recognize in myself oh so well! Concerning this problem, I would have emphasized at the end (as a valuable conclusion) that the maximum area is always reached with an isosceles triangle, and is therefore half the area of the imaginary square with (equal) sides 5/sqrt(2) and fixed diagonal 5. For example, if you take a fixed perimeter for rectangles in different side lengths, you will reach the maximum area with a square with the same perimeter, and thus for isosceles triangles the same... Thank you Newton and stay well, Jan-W p.s. Also think of constraints and objective function...
Sometimes I think I could add and subtract as quickly as I could but it doesn't happen. You are correct, emphasizing the general idea of maximum area being half the imaginary square would have been a quick tip even from the beginning since the largest rectangle is always a square.
@@PrimeNewtons ... I blame my uninspiring elementary school teacher (lol); how could I be responsible at that age ...
omg, u have uploaded this at the very right time!!!!!
I'm glad it's useful at this tome to you. Please share. Thank you.
@@PrimeNewtons I already shared it with my friends. I suggest please upload more optimization problems because a lot of students are having a hard time with this particular lesson. Thanks a lot!!
Largest area possible: for x=y. 1 + 1 = √2. So 5/√2
Very good. Thanks 👍
This is a fun puzzle if you don't over-think it. I was able to solve in 30 sec by knowing the largest rectangle bounded by its diagonal is formed by equal side lengths. The side length of such a square is the diagonal over root2. Multiply the side lengths and take half to get the largest triangle.
Taking this a bit further, the area is 1/2 x*y, but x=y. So area = 1/2 (x^2). Also, since it's an isocelese triangle x^2 = 1/2 (hyp)^2. So we get area = 1/2*(1/2*(5cm)^2) = 25/4
It is! Have you proved it? I am very interested in the relationship between symmetry and area. As you say, the largest area you can get in a four-sided figure, for a fixed perimeter, or a fixed diagonal, it seems, is a square, which is the most symmetric quadrilateral. I wonder how far this can be extended.
This is so helpful thank you
I had a different approach. Given that for the same hypotenuse, 45° 45° 90° triangle will have the largest area. By applying the property of the special triangle, I can do (5/√2)^2/2, that will be the product of the length and width/2, which is equal to 6.25
To find x, we may consider A^2=x^2(25-x^2)*1/4 or 4A^2 instead of A.
Hypotenuse^2=4×TriangleArea+(LegsDifference)^2
What does the 2nd derivative do? Can that be used to show we actually found the max?
The algebra is the hard part!
Where is your algebra course?
Check out my precalculus Playlist
@@PrimeNewtons Will do. Many thanks! 😃
thanks Newton
You're welcome
It is easier to optimize the square of the area
Good point, as we're looking for the maximum value of an expression on a positive interval, meaning the squared expression will reach its maximum value at the same x-coordinate.
why did you equate the denominator of the first derivative to zero when you were trying to find the CNs. The values of x which make the denominator zero are the ones which make it undefined, correct? so why didnt you just equate the numerator
fuck i love you, best vid possible
Triangle is isosceles
"WTF" 😂
Вы ЧО, с Урала!?!
1)maxA треугольника при maxh;
2) maxh=5/2;
3) maxA= 1/2 * 5 * 5/2 = 25/4;
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