# find the least possible value of x^2+y^2 given that x+y=10

I know the anwer, but this question has something to do with differentiation. I don't know how to use that to work out the answer?

Posted Wed 14th November, 2012 @ 22:20 by SS

Is it the root of 10?

Answered Tue 20th November, 2012 @ 16:32 by Chlo

Or 100?

Answered Tue 20th November, 2012 @ 16:32 by Chlo

its x=5 and y=5 but i can't explain why :(

it might be trigonmetry/ pythagoras problem because if you think of a triangle with sides x and y and hypoteneuse then when the angles are 45 degrees then it will give you the smallest hypotenuse and that can only happen if x=y so x=5 and y=5.

sin(45)=5/hypotenuse. so hypotenuse is 5 root 2 which multiplied by cos 45=5/ hypotenuse = 50

any other combination is higher by the looks of it.

im not sure whether this is the right way to do it but it kinda works and helps to explain it :/

either way hope it helps and inspires you to sum it up better than i can.

regards

RTF

Answered Tue 20th November, 2012 @ 16:50 by Ragnaros the Firelord
Edited by Ragnaros the Firelord on Tue 20th November, 2012 @ 17:04

and on another note, i think there is a number of methods to work this question out but solving by differentiation has me stumped. it may be solved by equation of the circle :/ but again im really unsure on how to do it.

Answered Tue 20th November, 2012 @ 17:12 by Ragnaros the Firelord

I don't have a clue if the following is right, but this is what I tried...

So if you differentiate the x^2 + y^2 then you will get
dx/dy = 2x + 2y

Now, if you want to find the minimum that they can be you need to find the point at which they add to make 0

0= 2x+ 2y
Therefore:
-2x = 2y

So then you know that one equals the other. The only possible way that the x and y can be the same and total to make 10 is if they are both 5. So your answer is 50 because 5^2 + -5^2 = 50 ...

Though the negative sign confuses me and I'm really not sure this is how you do it... Would be interested to hear how you finally did it though!

Thanks,
Tom

Answered Tue 27th November, 2012 @ 18:58 by Tom