find the smallest possible value of the sum of the squares
........2..2
S = X + Y
the above example is : s = x square + y square
2007-09-11 07:42:54 · 6 answers · asked by M R 1 in Science & Mathematics ➔ Mathematics
4² + 4² = 32
The others :
5² + 3² = 34
6² + 2² = 40
7² + 1² = 50
8² + 0² = 64
2007-09-11 07:46:17 · answer #1 · answered by antone_fo 4 · 2⤊ 0⤋
Figure out what numbers would add up to 8. So, there is
8+0
7+1
6+2
5+3
4+4
Of these, 4+4 would generate the smallest sum of the squares = 4^2 + 4^2 = 16 + 16 = 32
If you square any other number, you will get a higher sum of the squares than 32 (5+3 is next smallest sum -- 5^2 + 3^2 = 25+9 = 34).
2007-09-11 07:50:17 · answer #2 · answered by jemt113 2 · 0⤊ 0⤋
x+y=8
S=x^2+y^2
=x^2+y^2+2xy-2xy
=(x+y)^2-2xy
=8^2-2x(8-x)
=64-16x+2x^2
If this is calculus level, you can take the derivative and set that equal to 0. But I'm guessing it's an algebra level problem.
This is parabola that's concave up. The lowest value is the vertex, with x = -b/2a = -(-16)/(2*2) = 4
S(4)=64-16*4+2*4^2
=32
2007-09-11 07:53:06 · answer #3 · answered by np_rt 4 · 0⤊ 0⤋
4, 4 -> 32
2007-09-11 07:48:56 · answer #4 · answered by gebobs 6 · 0⤊ 0⤋
uhhhhhhh
2007-09-11 07:49:46 · answer #5 · answered by Mohsy 3 · 0⤊ 0⤋
Hmmm, that is SEXY...
2007-09-11 07:49:13 · answer #6 · answered by geezuskreyest 5 · 0⤊ 0⤋