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 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.
@@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!!
....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.
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
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
![How to Solve ANY Optimization Problem [Calc 1]](http://i.ytimg.com/vi/cUMvwG7wvzM/mqdefault.jpg)
You are my favourite math tutor on youtube
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 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.
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!!
This is so helpful thank you
Very good. Thanks 👍
....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 ...
To find x, we may consider A^2=x^2(25-x^2)*1/4 or 4A^2 instead of A.
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
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
What does the 2nd derivative do? Can that be used to show we actually found the max?
Hypotenuse^2=4×TriangleArea+(LegsDifference)^2
It is easier to optimize the square of the area
thanks Newton
You're welcome
fuck i love you, best vid possible
The algebra is the hard part!
Where is your algebra course?
Check out my precalculus Playlist
@@PrimeNewtons Will do. Many thanks! 😃
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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