The minimum value of $(x - 6)^{2} + (x + 3)^{2} + (x - 8)^{2} + (x + 4)^{2} + (x - 3)^{2}$ is?

114

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141

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104

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2

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Solution

The correct option is A $114$
$= (x - 6)^{2} + (x + 3)^{2} + (x - 8)^{2} + (x + 4)^{2} + (x - 3)^{2}$
$= x^{2} + 36 - 12 x + x^{2} + 9 + 6 x + x^{2} + 64 - 16 x + x^{2} + 16 + 8 x + x^{2} + 9 - 6 x$
$= 5 x^{2} - 34 x + 14 x + 134$
$= 5 x^{2} - 20 x + 134$
$f (x) = 5 x^{2} - 20 x + 134 a = 5, b = - 20, c = 134 a > 0$ (minimum point)
$f (x) = 6 x - 20$ maximum point $= (\frac{- b}{2 a}, f (\frac{- b}{2 a}))$
$x = \frac{+ 20}{10} x = 2$
$y = 5 (2)^{2} = 20 (2) + 134 = 20 - 40 + 134$
min value $= 114$

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