Question

The sum of two nonnegative numbers is 20. Find the numbers if the sum of their squares is as large as possible; as small as possible.

Answer 1

as large as possible you do in your head
since you cannot tell the two apart, largest is when they are equal (by symmetry)

Answer 2

you can use algebra if you like i suppose
make your teacher happy maybe
find the minimum of \(x^2+(20-x)^2\)

Answer 3

man i got that backwards
min you do in your head not max

Answer 4

i apologize, but i forgot how to do maxes and mins....like, how do you find the different combinations?

Answer 5

for min they are both 10, for max one is 20, the other is 0

Answer 6

one number is \(x\) say, then since the total is 20, the other must be \(20-x\)
sum of the squares is
\[x^2+(20-x)^2=2x^2-40x+400\] parabola that face up

Answer 7

minimum is at the vertex
first coordinate of the vertex is \(-\frac{b}{2a}=-\frac{-40}{4}=10\)

Answer 8

max is at the endpoint of the domain
domain is \([0,20]\) and this is entirely symmetric about 10, so max is where one is 20 and the other is zero

Answer 9

Oh ok, thanks so much!

Answer 10

yw