Find Minimal Sum of Distances: (1,2) & (4,3) on Axis OX

[leprofece]

find a point on the axis OX whose sum of distances to landmarks: (1, 2) and (4, 3) is minimal.

Answer (2,0)

(x-1)₂+(y-2)^2 +
(x-4)₂+(y-3)^2 = D

Y = mx
I don't know if solve each distance apart or how i wrote??

 

[MarkFL]

Re: sum of distances

Your point on the $x$-axis may be written as $(x,0)$, and so the objective function is:

\(\displaystyle f(x)=\sqrt{(x-1)^2+4}+\sqrt{(x-4)^2+9}\)

Minimizing this gives a different result than what you cite though. (Thinking)

 

[leprofece]

Re: sum of distances

I don't know but the 2 roots must be
((x-2)²+4)^1/2+ (x²+9)^1/2 = D When I derive an equated to 0 I got (6,0)
the roots that my friend markflo wrote if he didnot make a mistake why are they so??

 

[HallsofIvy]

Re: sum of distances

No, the answer is NOT (2, 0).

The simplest way to do this problem is:

Change the second point to (4, -3) on the opposite side of the x-axis. A straight line is the shortest distance between two point and the straight line from (1, 2) to (4, -3), which has equation y= (-5/3)x+ 11/3, crosses the x-axis when (-5/3)x+11/3= 0 or x= 11/5= 2.2.

Now, draw a picture showing all three points, (1, 2), (4, 3), and (4, -3) and use "congruent triangles" to show that the answer to the given problem is also (2.2, 0).

(This is referred to as the "method of reflections.)

 

[leprofece]

The real answer is (21/5,0)
do it Change the way of solution?

 

[MarkFL]

The real answer is (21/5,0)
do it Change the way of solution?

No, the point is \(\displaystyle \left(\frac{11}{5},0 \right)\).

You can find this either minimizing the funtion I gave or much more simply using the method of reflection as shown by HallsofIvy.

 

[leprofece]

No, the point is \(\displaystyle \left(\frac{11}{5},0 \right)\).

You can find this either minimizing the funtion I gave or much more simply using the method of reflection as shown by HallsofIvy.

Your point on the x-axis may be written as (x,0), and so the objective function is:

f(x)=(x−1)2+4−−−−−−−−−−√+(x−4)2+9−−−−−−−−−−√

ok friend mark Thanks
I MUST minimize because I an asked to do so
I tried deriving and I did not get it
OK
operating
(x²-2x+5)^1/2+(x²-8x+25)^1/2

Deriving
(x-1)/(x²-2x+5)^1/2+(x-4)/(x²-8x+25)^1/2

Equating to 0
(x-4)(x²-2x+5)^1/2+(x-1)(x²-8x+25)^1/2= 0

And Squaring
(x-4)²(x²-2x+5) =(x-1)²(x²-8x+25)

I don't know here if I and doing well because I get a cubic equation after that

Please tell and help me (Blush)

 

[MarkFL]

Okay, our objective function is:

\(\displaystyle f(x)=\sqrt{(x-1)^2+4}+\sqrt{(x-4)^2+9}\)

So, differentiating and equating the result to zero, we obtain:

\(\displaystyle f'(x)=\frac{x-1}{\sqrt{(x-1)^2+4}}+\frac{x-4}{\sqrt{(x-4)^2+9}}=\frac{(x-1)\sqrt{(x-4)^2+9}+(x-4)\sqrt{(x-1)^2+4}}{\sqrt{(x-1)^2+4}\sqrt{(x-4)^2+9}}=0\)

We see there are no real roots in the denominator, so we are left with:

\(\displaystyle (x-1)\sqrt{(x-4)^2+9}+(x-4)\sqrt{(x-1)^2+4}=0\)

\(\displaystyle (x-1)\sqrt{(x-4)^2+9}=(4-x)\sqrt{(x-1)^2+4}=0\)

Square both sides:

\(\displaystyle (x-1)^2\left((x-4)^2+9 \right)=(x-4)^2\left((x-1)^2+4 \right)\)

Distribute:

\(\displaystyle (x-1)^2(x-4)^2+9(x-1)^2=(x-1)^2(x-4)^2+4(x-4)^2\)

Arrange as:

\(\displaystyle 9(x-1)^2-4(x-4)^2=0\)

Factor as difference of squares:

\(\displaystyle \left(3(x-1)+2(x-4) \right)\left(3(x-1)-2(x-4) \right)=0\)

Distribute and collect like terms:

\(\displaystyle (5x-11)(x+5)=0\)

Checking both roots, we find the root $x=-5$ is extraneous, but the root \(\displaystyle x=\frac{11}{5}\) is valid.

Now, to show that this critical value is at a minimum, we can use the first derivative test:

\(\displaystyle f'(1)=-\frac{1}{\sqrt{2}}<0\)

\(\displaystyle f'(4)=\frac{3}{\sqrt{13}}>0\)

and so we may conclude that the point:

\(\displaystyle \left(\frac{11}{5},0 \right)\)

is the solution.

 

[leprofece]

The only thing that i don't know
Is where do they some values come from??
f(x)=(x−1)²+4−−−−−−−−−−√+(x−4)²+9−−−−−−−−−−√

One Point is (1,2)
another is (4,3)
I understood (x - 1²)²+2² This is root 1
The second root = (x - 2²)²+3²

I hope It will be so?

 

[MarkFL]

The distance from the variable point $(x,0)$ and the given fixed point $(1,2)$ is:

\(\displaystyle \sqrt{(x-1)^2+(0-2)^2}=\sqrt{(x-1)^2+4}\)

The distance from the variable point $(x,0)$ and the given fixed point $(4,3)$ is:

\(\displaystyle \sqrt{(x-4)^2+(0-3)^2}=\sqrt{(x-4)^2+9}\)

The sum of these distances is our objective function:

\(\displaystyle f(x)=\sqrt{(x-1)^2+4}+\sqrt{(x-4)^2+9}\)