Expert: Paul Klarreich - 4/4/2008

Question
Prove that the optimum value of y=((x-a)^2)+((x-b)^2) occurs when x=(a+b)/2. I have no idea where to start. Can you help me get started?  The optimum is either a max or a minimum value of.  I think you have to convert it into a quadratic equation and maybe complete the square to find the vertex??

Answer
Questioner:   Patrick
Category:  Advanced Math
Private:  No
 
Subject:  Quadratics
Question:  Prove that the optimum value of y=((x-a)^2)+((x-b)^2) occurs when x=(a+b)/2. I have no idea where to start. Can you help me get started?  The optimum is either a max or a minimum value of.  

>> Of what?

I think you have to convert it into a quadratic equation

>> quadratic FUNCTION.  And you don't have to convert it; it already IS a quadratic function.

and maybe complete the square to find the vertex??
............................................  
Hi, Patrick,

Yes, that sounds like a good idea. (in this case, no completing will be needed)  The graph of

y = px^2 + qx + r

has a vertex at  x = -q/2p.  So all you have to do is manipulate your function into this form, match things up, and go.

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

y = x^2 - 2ax + a^2 + x^2 - 2bx + b^2

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

Now p = 2,  q = - 2a - 2b.

And your vertex (where you get your MINIMUM) is at
   -(-2a - 2b)
x = ------------
       2(2)

   2a + 2b
x = -------
    2(2)
 
   a + b
x = -----
     2

That's what you had to prove.