document.write( "Question 57616: 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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Algebra.Com's Answer #39354 by venugopalramana(3286) $\"\"$ $\"About$

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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?\r
\n" ); document.write( "\n" ); document.write( "let the 2 odd numbers be 2n-1,2n+1
\n" ); document.write( "let the third number =x
\n" ); document.write( "sum = 2n-1+2n+1+x=4n+x=48
\n" ); document.write( "4n+x = 48.............i
\n" ); document.write( "sum of squares = s = (2n-1)^2 + (2n+1)^2+x^2= 4n^2+2+x^2
\n" ); document.write( "s = 4n^2+2+(48-4n)^2 = 20n^2-384n+2304+2=20n^2 - 384n + 2306
\n" ); document.write( "s = 20[n^2-19.2n+115.3]= 20[n-9.6]^2+462.8
\n" ); document.write( "hence s minimum when n=9.6 ...that is n=9 let us test
\n" ); document.write( "2*9-1=17 &19 &12......sum of sdquares = 17^2+19^2+12^2=794
\n" ); document.write( "is minimum \n" ); document.write( "

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