Let y=(x-a)^2+(x-b)^2, a, b constants. For what value of x is y a minimum?

1964 AMC 12 Problem 24

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24.

Let

y=(x-a)^2+(x-b)^2,

a,

b

constants. For what value of

x

y

a minimum?

\dfrac{a+b}{2}

a+b

\sqrt{ab}

\sqrt{\dfrac{a^2+b^2}{2}}

\dfrac{a+b}{2ab}

Expand and collect the terms in

x

Complete the square or use the vertex formula

Expanding and completing the square gives

y=2\left(x-\frac{a+b}{2}\right)^2+\frac{(a-b)^2}{2}.

The squared term is minimized at

x=\frac{a+b}{2}.

Therefore, the correct answer is A.