SOLUTION: two consecutive odd numbers and a third number add up to 48. what are the numbers so that the sum of their squares is a minimum?
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-> SOLUTION: two consecutive odd numbers and a third number add up to 48. what are the numbers so that the sum of their squares is a minimum?
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You can put this solution on YOUR website! two consecutive odd numbers and a third number add up to 48. what are the numbers so that the sum of their squares is a minimum?
let the 2 odd numbers be 2n-1,2n+1
let the third number =x
sum = 2n-1+2n+1+x=4n+x=48
4n+x = 48.............i
sum of squares = s = (2n-1)^2 + (2n+1)^2+x^2= 4n^2+2+x^2
s = 4n^2+2+(48-4n)^2 = 20n^2-384n+2304+2=20n^2 - 384n + 2306
s = 20[n^2-19.2n+115.3]= 20[n-9.6]^2+462.8
hence s minimum when n=9.6 ...that is n=9 let us test
2*9-1=17 &19 &12......sum of sdquares = 17^2+19^2+12^2=794
is minimum
