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Divide 12 into 2 parts such that sum of there squares is maximum |
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Answer» Let x = one of the numbersThen the other number = 12 - xSum of the squares is 74:x^2 + (12-x)^2 = 74Solve for x:x^2 + 144 - 24x + x^2 = 742x^2 - 24x + 70 = 0x^2 - 12x + 35 = 0Factor:(x-7)(x-5) = 0So the numbers are 5 and 7 |
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