Optimization- Maximum Area of Right Triangle with constant Hypotenuse.

In this video I showed how find the maximum area of right triangle with constant hypotenuse

[alanmuston]

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.

[Jadem]

You are my favourite math tutor on youtube

[alanmuston]

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-]

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-willemreens]

....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...

[aaliyahramos]

omg, u have uploaded this at the very right

[ilyayana-clnc]

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.

[AubreyForever]

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

[natureandus]

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

[oreoforlife]

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

[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.

[punditgi]

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

[aaronhoogendijk]

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

[souverainer]

It is easier to optimize the square of the area

[Бывалый-фб]

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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