# Homework Help: Proof for Max/Min Problem

1. Mar 8, 2006

### ohlhauc1

We had to do a question in my Advanced Mathematics class, and the way that I did the problem was supposedly right. My teacher even did it that way. However, the answer was wrong, so my teacher showed us the correct way to do the problem.

The dilemma I have is that BOTH ways should work, but they don't. :surprised Therefore, I am going to give both solutions to the problem, and I was wondering if you could tell me why the first one does not work.

Question: The coordinates of points P and Q are (1,2) and (2,-3), respectively, and R is a point on the line x=-1. Find the coordinates of R so that PR + RQ is a minimum?

Here is the graph/image I used for this problem:
http://s62.yousendit.com/d.aspx?id=0LUTJESKXM0R33IOBSY2IH1Z3A

MY SOLUTION:

Equations: PR^2 = 2^2 + y^2
QR^2 = 3^2 + (5 - y)^2

PR^2 + QR^2 = minimum or m
2^2 + y^2 + 3^2 + (5 -y)(5 - y) = minimum
4 + y^2 + 9 + 25 - 10y + y^2 = minimum
2y^2 - 10y + 38 = minimum

y = -b/2a = 10/4 = 5/2

Y-coordinate of R = 2 - 5/2 = -1/2

Therefore, the coordinates of R are (-1, -1/2)

CORRECT SOLUTION:

Let m = minimum

mP'R = [ (2 - y) / (-3 + 1) ] = [ (2 - y) / -2 ]

mRQ = [ (y + 3) / (-1 -2) ] = [ (y + 3) / -3 ]

Note: mP'R = mRQ
[ (2 - y) / -2 ] = [ (y + 3) / -3 ]
= -2y - 6 = 3y - 6
= -5y = 0
y = 0

The coordinates of R are therefore (-1, 0).

*So...I have presented both solutions, and I would highly appreciate your help. I am really interested in trying to understand this dilemma.

Thanks

2. Mar 9, 2006

### NateTG

The maximum of the sum of the squares isn't necessarily the maximum of the sum.