Why Does One Mathematical Approach Fail in This Optimization Problem?

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Therefore, in summary, the coordinates of point R can be either (-1, -1/2) or (-1, 0) in order to minimize the sum of distances PR and RQ. The first solution is based on the equation of a circle and the second solution is based on the slopes of the line segments. Both solutions have their own reasoning and are correct, but it is unclear why the first solution did not work.
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ohlhauc1
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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. 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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  • #2
The maximum of the sum of the squares isn't necessarily the maximum of the sum.
 
  • #3
for sharing your question and dilemma about this problem in your Advanced Mathematics class. I am happy to provide my response and help you understand this dilemma.

First of all, it is important to note that there can be multiple ways to solve a problem, and both of your solutions are valid approaches. However, the key to finding the correct solution is to carefully consider the given information and use the appropriate mathematical concepts and equations.

In your first solution, you used the distance formula to find the minimum value of PR^2 + QR^2. However, this approach assumes that the point R can be any point on the line x = -1, which is not true. The problem states that R is a point on the line x = -1, which means that its x-coordinate is always -1, and only the y-coordinate can vary. Therefore, your solution does not take into account the constraint that R must lie on the line x = -1.

In the correct solution, the concept of minimizing the distance between two points is used. This approach considers the given information that R must lie on the line x = -1, and uses the slope of the line to find the y-coordinate of R that minimizes the distance between P and R, and between R and Q. This approach takes into account the constraint of R being on the line x = -1 and gives the correct solution.

In conclusion, the key to finding the correct solution to this problem is to carefully consider the given information and use the appropriate mathematical concepts. Both of your solutions are valid approaches, but the second one takes into account the given constraint and gives the correct answer. I hope this helps you understand the dilemma and provides some clarity on why the first solution did not work. Keep up the good work in your Advanced Mathematics class!
 

FAQ: Why Does One Mathematical Approach Fail in This Optimization Problem?

What is the Proof for Max/Min Problem?

The Proof for Max/Min Problem is a method used to find the maximum or minimum value of a function. It involves taking the derivative of the function and setting it equal to zero to find the critical points, then plugging these points into the original function to determine which point gives the maximum or minimum value.

Why is the Proof for Max/Min Problem important?

The Proof for Max/Min Problem is important because it allows us to find the optimal solution for a given function. This is useful in various fields such as economics, physics, engineering, and more. It also helps us to understand the behavior of a function and its critical points.

What are the steps involved in the Proof for Max/Min Problem?

The steps involved in the Proof for Max/Min Problem are: 1. Take the derivative of the function 2. Set the derivative equal to zero and solve for the critical points 3. Determine the second derivative of the function 4. Use the second derivative test to determine the maximum or minimum value

Can the Proof for Max/Min Problem be applied to any function?

Yes, the Proof for Max/Min Problem can be applied to any differentiable function. However, it may not always give a definitive answer as there may be multiple critical points or the function may not have a maximum or minimum value.

Are there any limitations to the Proof for Max/Min Problem?

One limitation of the Proof for Max/Min Problem is that it only gives local maximum or minimum values. It does not guarantee that the found point is the absolute maximum or minimum for the entire function. Additionally, the function must be differentiable for this method to be applied.

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