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Is this possible to solve without partial derivatives? in this i asked about to find minima of the function $f(x,y)=\sqrt{x^2-14x+74}+ \sqrt{y^2-4y+20}+ \sqrt{x^2+y^2-10x-10y+50}$, this is my progress by considering two fixed points in cartesian plane , $A(1,2)$ and $B(7,0)$ while the rest ones $I(x,5)$ and $F(5,y)$ are varying , now problem becomes to minimize $AF + FI +BI$ , for this i considered reflection of B about line $x=5$ , now minimum distance would be a straight line between $(7,10)$ and $(1,2)$ but when i put into original funtion it doesnt give minimum to be $10$ , can anyone tell how to correct it and get the required answer of $5+\sqrt{29}$ doing this way only?enter image description here

Paracetamol
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2 Answers2

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To have notations that i can follow in a simpler manner, let me introduce the points similar to the ones in the OP, the idea from the OP is almost all we need: $$ \begin{aligned} A &= (2,1)\ ,\\ B &= (10,7)\ ,\\ C &=(5,5)\ ,\\[2mm] X &= (x,5)\ ,\\ Y &= (5, y)\ , \end{aligned} $$ and let us also consider the points $X'$, $Y'$ obtained by reflecting $X,Y$ w.r.t. the point $C$. The picture is as follows:

MSE 4237303

Then the length of the segments $AX$, then $XY$, then $YB$ are respectively the square roots of $(x-2)^2 + (5-1)^2$, then $(x-5)^2 +(y-5)^2$, and $(y-t)^2+5^2$, so we have to minimize $AX+XY+YB$.

A few words, addressing the question about what is wrong in the OP with taking the length of $AB$. (My $B$ is a corrected version.)

Unfortunately, the segment $AB$ cannot be split in three segments to realize this minimum without overlap, since drawing the segment from A to B we first cut the vertical line where $Y$ lives on, but we have to take first the $X$ point in the sum $AX+XY+YB$... (Trying to realize the segment as a sum we travel from $A$ to that point of intersection on the horizontal line, than we come back to the vertical line, than we go to $B$. So this would be an other problem... minimizing $AY+YX+XB$, which is an other function.)

We divide the problem in four cases, so that $X,X';Y,Y'$ run strictly on the rays from $C$ as shown in the picture. (So $x\le 5$, $x'=10-x\ge 5$; $y\le 5$, $y'=10-y\ge 5$.) Note that $XYX'Y'$ is a rhombus, its sides are equal.

The cases lead to the four values of the function $f$ corresponding to $$ \begin{aligned} E &= AX+XY+YB\ ,\\ E' &= AX+XY'+Y'B\ ,\\ F &= AX'+X'Y+YB\ ,\\ F' &= AX'+X'Y'+Y'B\ .\\[3mm] &\qquad\text{Observe now that...}\\ AX+XY &= AX+XY'\ ,\\ AX'+X'Y &= AX'+X'Y'\ ,\\ XY+YB &= X'Y+YB\ ,\\ XY'+Y'B &= X'Y'+Y'B\ ,\\ &\qquad\text{so adding one more term...}\\ E = AX+XY+YB &= AX+XY'+YB \color{red}{\ge} AX+XY'+Y'B =E'\ ,\\ F = AX'+X'Y +YB &= AX'+X'Y' +YB \color{red}{\ge} AX'+X'Y'+Y'B =F'\ ,\\ E = AX + XY+YB &= AX + X'Y+YB \color{blue}{\le} AX' + X'Y+YB = F\ ,\\ F = AX + XY'+Y'B &= AX + X'Y'+Y'B\color{blue}{\le} AX' + X'Y+Y'B = F'\ . \end{aligned} $$ So the minimal value is obtained for $E'$, i.e. for the positions of $X$ and $Y'$ in the picture. Now we move only these two points. Fixing $X$, the minimal value is obtained for $Y'$ on the line $XB$, so $AX+XY'+Y'B=AX+XB$ has to be minimized now. It is clear that the minimal value is reached for $X$ in $C$ (when $Y'$ is also in $C$).

dan_fulea
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  • So my mistake was instead of choosing A to be (1,2) i should have taken A to be (-2,-1) in that way i could have split the line segment by reflection technique two times instead of one i did intially so that we get that line A(-2,-1)B (7,10) to divided into three segments? – Paracetamol Sep 01 '21 at 15:22
  • And i have thought of another method (vector) can u once check if that is helpful too in getting the minima ? I will post in the comments only – Paracetamol Sep 01 '21 at 15:27
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    After i typed and have the drawing, i saw that my points $A(2,1)$ and $B(10,7)$ correspond to the reflected points from the OP $A(1,2)$ and $B'(7,10)$. Now the problem is the following using your notations. Yes, $AB'=\sqrt{(7-1)^2+(10-2)^2}=10$, but you cannot realize this $10$, segment length, as the sum of three segments without overlapping which are in order $AF$ (with $F$ on the vertical line through $C(5,5)$) then $FI$ (with $I$ on the horizontal line through $C$) and then $IB'$. Your ray $AB'$ starting in $A$ first hits the horizontal, after this the vertical through $C$. – dan_fulea Sep 01 '21 at 15:34
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    (In fact, the OP does need $B$, only $B'$ is enough to expose the idea and present it in a minimal form.) I cannot see how to use the point $(-2,-1)$ instead of $(1,2)$ as a geometric aid. The above long answer could have been given as a picture answer without any comments. Just consider the four cases from the picture, only one case is minimal, and in this case the minimal value is taken in $C$... – dan_fulea Sep 01 '21 at 15:38
  • Thanks got it . – Paracetamol Sep 01 '21 at 17:42
  • One more point how will one reason to remove the point A(2,9 ) and B(0,7) from consideration ? – Paracetamol Sep 02 '21 at 07:59
  • I think it will not affect anyhow , because those will be just reflections along original A and B in your sol about y = 5 and x=5 right ? – Paracetamol Sep 02 '21 at 15:34
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$f(x,y)$ can be writen as

$$f(x,y)=\sqrt{(x-7)^2+25}+\sqrt{(y-2)^2+16}+\sqrt{(x-5)^2+(y-5)^2}$$

To simplify we replace the variables:

$u=x-5\rightarrow x=u+5$

$v=y-5\rightarrow y=v+5$

and the function becomes

$$g(u,v)=\sqrt{(u-2)^2+25}+\sqrt{(v+3)^2+16}+\sqrt{u^2+v^2}$$

Now the function is the sum of distances between the points:

$d_1(A(2,\pm5), B(u,0))=\sqrt{(u-2)^2+25}$

$d_2(B(u,0), C(0,v))=\sqrt{u^2+v^2}$

$d_3(C(0,v),D(\pm4,-3))=\sqrt{(v+3)^2+16}$

For $A(2,5); B(-4,-3)$ the minimum of $g(u,v)$ would be achieved when these 4 points are colinear, hence the points $B$ and $C$ are on the line through $A$ and $D$.

The equation of the line is

$$\frac{v+3}{u+4}=\frac{4}{3}$$

This line intersects the axes in the points $(0, 7/3)$ and $(-7/4, 0)$ so the values of $u_m,v_m$ are $-7/4, 7/3$.

Now, $$min(f(x,y))=min(g(u,v))=g(-7/4, 7/3)=25/4+20/3+35/12=95/6$$

And the length is bigger than $5+\sqrt{29}$!

Where is the mistake?

The mistake is that the segment $CD$ is scanned twice!

So the minimum length is achieved only when

$|CD|=0\rightarrow u=0, v=0\rightarrow L_{min}=5+\sqrt{29}$.

Vasile
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