Question 194616: one number is 30 less that twice another number. find the 2 numbers if the sum of their squares is a minimum.
Answer by RAY100(1637)   (Show Source):
You can put this solution on YOUR website! let n=2m -30
and n^2 + m^2 = min
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(2m-30)^2 +m^2 =0
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4m^2 -120m +90 +m^2 =0
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3m^2 -120 m +90 =0
3(m^2 -40m +30 ) =0
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use quadratic eqn , a=3, b=-40,c=30
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m= ( -(-40) +/- sqrt( -(-40) - 4 (1)(30) ) / 2(1)
m = (40 +/- sqrt (1600-120) )/2
m= (40 +/- 38.47) /2 = 39.2 ,,,, or .765
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subst for n,
n= 2m-30 = 2*39.2 -30 = 48.4
or n= 2* .765 -30 = 28.47
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pairs are m= 39.2, n=48.4
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,,,,,,,,m=.765, n= 28.47
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problem asks that sum of squares is a minimum, therefore use, m=.765, n=28.47
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