# What is the mistake in my approach and how to rectify it?

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?

• Did you mean reflection of B about line y = 5 and not x = 5? Commented Aug 31, 2021 at 11:24
• Yeah {}{}{}{}{} Commented Aug 31, 2021 at 11:26
• I think problem is that AI and IB' do not lie on the same line. Commented Aug 31, 2021 at 11:27

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:

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 $$C$$).

• 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? Commented Sep 1, 2021 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 Commented Sep 1, 2021 at 15:27
• 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 $C$. Commented Sep 1, 2021 at 15:34
• (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 $C$... Commented Sep 1, 2021 at 15:38
• Thanks got it . Commented Sep 1, 2021 at 17:42

$$f(x,y)$$ 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

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

• Gud approach thanks Commented Sep 1, 2021 at 17:42