Optimization- Maximum Area of Right Triangle with constant Hypotenuse.

[@alanmuston3554]

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.

[@Jadem4]

You are my favourite math tutor on youtube

[@alanmuston3554]

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!

[@-opresiet-1414]

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.

[@jan-willemreens9010]

....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...

[@PrimeNewtons]

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.

[@jan-willemreens9010]

@@PrimeNewtons ... I blame my uninspiring elementary school teacher (lol); how could I be responsible at that age ...

[@surendrakverma555]

Very good. Thanks 👍

[@aaliyahramos5033]

omg, u have uploaded this at the very right time!!!!!

[@PrimeNewtons]

I'm glad it's useful at this tome to you. Please share. Thank you.

[@charlizeaaliyahramos]

@@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!!

[@zuchilochikee5727]

This is so helpful thank you

[@natureandus6565]

To find x, we may consider A^2=x^2(25-x^2)*1/4 or 4A^2 instead of A.

[@ilyayana-cl4nc]

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.

[@oreoforlife720]

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

[@AubreyForever]

What does the 2nd derivative do? Can that be used to show we actually found the max?

[@flavrt]

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.

[@mikefochtman7164]

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

[@alanmuston3554]

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.

[@punditgi]

The algebra is the hard part! Where is your algebra course?

[@PrimeNewtons]

Check out my precalculus Playlist

[@punditgi]

@@PrimeNewtons Will do. Many thanks! 😃

[@elliotappiah7434]

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

[@evidenciachengeta6572]

thanks Newton

[@PrimeNewtons]

You're welcome

[@aaronhoogendijk239]

Largest area possible: for x=y. 1 + 1 = √2. So 5/√2

[@fsyi8395]

Hypotenuse^2=4×TriangleArea+(LegsDifference)^2

[@souverain1er]

It is easier to optimize the square of the area

[@holyshit922]

Triangle is isosceles

[@leonvillemin6643]

fuck i love you, best vid possible

[@justarandomnerd3360]

"WTF" 😂

[@Бывалый-ф9б]

Вы ЧО, с Урала!?! 1)maxA треугольника при maxh; 2) maxh=5/2; 3) maxA= 1/2 * 5 * 5/2 = 25/4; ГЫЫЫЫЫЫЫЫЫЫЫЫЫЫ