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23 June, 04:04

Two numbers have a difference of 24. What is the sum of their squares if it is a minimum?

288

1156

366

576

+5
Answers (1)
  1. 23 June, 05:46
    0
    Let "a" and "b" be some number where:

    a - b = 24

    We want to find where a^2 + b^2 is a minimum. Instead of just logically figuring out that the answer is where a=b=12, I'll just use derivatives.

    So we can first substitute for "a" where a = b+24

    So we have (b+24) ^2 + b^2 = b^2 + 48b + 576 + b^2

    And that equals 2b^2 + 48b + 576

    Then we take the derivative and set it equal to zero:

    4b + 48 = 0

    4 (b+12) = 0

    b + 12 = 0

    b = - 12

    Thus "a" must equal 12.

    So:

    a = 12

    b = - 12

    And the sum of those two numbers squared is (12) ^2 + (-12) ^2 = 144 + 144 = 288.

    The smallest sum is 288.
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