# Can anyone help me to solve this problem. Thank.

• July 28th 2006, 02:20 AM
SuperTu
Can anyone help me to solve this problem. Thank.
please help me on the problem of finding minimum value. Thank.

http://www.math.um.edu.my/Ismweb/Ome.../Problem+1.pdf
• July 28th 2006, 04:12 AM
topsquark
Quote:

Originally Posted by SuperTu
please help me on the problem of finding minimum value. Thank.

http://www.math.um.edu.my/Ismweb/Ome.../Problem+1.pdf

$y = \sqrt{(x+a)^2+b} + \sqrt{(x+c)^2+d}$

I presume you mean the minimum value of y?

Set $\frac{dy}{dx} = \frac{2(x+a)}{\sqrt{(x+a)^2+b}} + \frac{2(x+c)}{\sqrt{(x+c)^2+d}} = 0$

Find a common denominator:
$\frac{2(x+a)\sqrt{(x+c)^2+d} + 2(x+c)\sqrt{(x+a)^2+b}}{\sqrt{(x+a)^2+b} \sqrt{(x+c)^2+d}}=0$

Thus:
$2(x+a)\sqrt{(x+c)^2+d} + 2(x+c)\sqrt{(x+a)^2+b} = 0$

$(x+a)\sqrt{(x+c)^2+d} = -(x+c)\sqrt{(x+a)^2+b}$

Square both sides:
$(x^2 + 2ax + a^2) (x^2 + 2cx + c^2 + d)$ = $(x^2 + 2cx + c^2) (x^2 + 2ax + a^2 + b)$

If you expand this on both sides of the equation you will find that you are left with a quadratic (there are a number of nice cancellations) that you can solve for x. Then put this value of x into the y equation to find a value of y. (Make sure that the final value you claim is the answer is, in fact, a minimum not a maximum.)

A quicker way to do this is logical. You are adding two positive terms to find a minimum, so find the minimum values of $\sqrt{(x+a)^2+b}$ and $\sqrt{(x+c)^2+d}$. Then add these to get the minimum value for y.

-Dan
• July 28th 2006, 07:17 AM
ThePerfectHacker
There is a much more simplified version.
$
y = \sqrt{(x+a)^2+b} + \sqrt{(x+c)^2+d}
$

If and only if,
$y=(x+a)^2+b+(x+c)^2+d$
(This is the same idea you use in the Method of Least squares).

Anyway, you can do it like topsquark said with the derivative but you can even do it without it.
Open,
$y=x^2+2ax+a^2+b+x^2+2cx+c^2+d$
Organize,
$y=2x^2+(2a+2c)x+(a^2+c^2+b+d)$
This is parabola with a minimum point which occurs at,
$x=-\frac{2a+2b}{4}=-1/2(a+b)$
• July 28th 2006, 08:16 PM
SuperTu
you all didn't see the question clearly
you all didn't see the question clearly.. there should be b square and d square in that equation but not b anf d. thank
• July 29th 2006, 04:37 AM
topsquark
Quote:

Originally Posted by SuperTu
you all didn't see the question clearly.. there should be b square and d square in that equation but not b anf d. thank

My apologies for reading it wrong, but the method of solution is identical.

-Dan
• July 29th 2006, 05:43 AM
SuperTu
may i know what theorem or statment should be used to explain that the stationary is minimum point instead of maximum or inflection point. thank
• July 29th 2006, 06:51 AM
topsquark
Quote:

Originally Posted by SuperTu
may i know what theorem or statment should be used to explain that the stationary is minimum point instead of maximum or inflection point. thank

In general there isn't one. Personally I would use the characteristics of the graph of the function to decide if we have a minimum point there, but in general you need to use the second derivative test:

If $\frac{d^2y}{dx^2}<0$ then the stationary point is a relative maximum.

If $\frac{d^2y}{dx^2}>0$ then the stationary point is a relative minimum.

If $\frac{d^2y}{dx^2}=0$ then the stationary point is a point of inflection.

I mentioned that we need to check this near the bottom of my first post.

-Dan
• July 29th 2006, 05:26 PM
ThePerfectHacker
Let me add something.
$\frac{d^2y}{dx^2}=0$--->Inflection point.
But you know nothing about whether it is maximum of minimum! You need to return back to the first derivative test. A common error is that people think whenever the second derivative is zero it means it is neither a minimum nor a maximum (I think the name of the term is 'saddle point')
• July 29th 2006, 09:23 PM
Soroban
Hello, Super!

Quote:

Given that $a,b,c,d$ are real numbers,

find the minimum value of: . $y\;=\;\sqrt{(x+a) + b^2} + \sqrt{(x+c)^2+d^2}$

Consider the following problem . . .

Given two points $Q(-a,b)$ and $R(-c,d)$, find the point $P(x,0)$ on the x-axis
. . such that the sum of the distances $\overline{PQ} + \overline{PR}$ is a minimum.
Code:

                R      |                 *(-c,d) |       Q        /        | (-a,b)*      /        |       \    /          |         \  /          |         \ /            |       - - * - - - - - - + - - -           P         (x,0)

Since $PQ\:=\:\sqrt{(x -[-a])^2 + (0 - b)^2} \:=\:\sqrt{(x + a)^2 + b^2}$

. and $PR\:=\:\sqrt{(x - [-c])^2 + (0 - d)^2} \;= \;\sqrt{(x + c)^2 + d^2}$

then: . $S\:=\:\sqrt{(x+a)^2 + b^2} + \sqrt{(x+c)^2 + d)^2}$

. . and we have a geometric interpretation of your problem.

• July 30th 2006, 01:31 AM
CaptainBlack
Will you cut us in for a share of the mystery gift if you win it?

I don't think you should be asking for our help, nor should we be giving it.

This link tells us about the context of this problem

RonL
• July 30th 2006, 10:26 AM
earboth
Quote:

Originally Posted by Soroban
...

Given two points $Q(-a,b)$ and $R(-c,d)$, find the point $P(x,0)$ on the x-axis
. . such that the sum of the distances $\overline{PQ} + \overline{PR}$ is a minimum.
Code:

                R      |                 *(-c,d) |       Q        /        | (-a,b)*      /        |       \    /          |         \  /          |         \ /            |       - - * - - - - - - + - - -           P         (x,0)
...

Hello, Soroban,

your explanation of the situation and especially your sketch tickled my memory: This problem was solved by Archimedes already - if I remember correctly. (Reflect R at the x-Axis. The straight line between Q and R' isthe shortest distance between Q and R. P is the zero of this line)

Greetings

EB