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.(adsbygoogle = window.adsbygoogle || []).push({});

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

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# Homework Help: Proof for Max/Min Problem

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