# Minimum value of the given function

Minimum value of $$\sqrt{2x^2+2x+1} +\sqrt{2x^2-10x+13}$$ is $$\sqrt{\alpha}$$ then $$\alpha$$ is________ .

Attempt

Wrote the equation as sum of distances from point A(-1,2) and point B(2,5) as $$\sqrt{(x+1)^2 +(x+2-2)^2} +\sqrt{(x-2)^2 + (x+2-5)^2}$$ Hence the point lies on the line y=x+2 it is the minimum sum of distance from the above given two points.

But from here I am not able to get the value of x and hence $$\alpha$$. Any suggestions?

• Too bad you haven't really read other solutions. Dec 2, 2018 at 17:06
• @greedoid Why you said that? Dec 2, 2018 at 18:41
• Because you upvote all solution imediatly after accepting first answer which does not tell you actualy how to find a solution and why to take tis substitution $t=x-1$... and last solution is not really a solution. Dec 2, 2018 at 18:58
• Okay. Understood your point. But, how come you can say that I didn't get the answers from another's method. In case of Robert, he used the concepts that I have been taught previously. In case of Cesareo,Boshu and farruhota, the method is almost same and the same thinking I have used. Also in your case, you have also used geometric approach but by triangle inequality taking another point. But now for selecting answer I would upvote the answer which I think gave another view also. Hope it clears all the "fogs". BTW thanks for suggesting me another method. Dec 2, 2018 at 19:06

There are a number of ways to do this, including brute force and calculus, but since you already found out the rather nice geometric interpretation as the sum of the distances from the given points, let's do that.

A few things will come in handy here.

1. The line on which $$A$$ and $$B$$ lie is $$y=x+3$$, which is parallel to your line of interest (i.e. $$y=x+2$$).
2. Both lines have slope $$1$$.

Here's the general idea: Note that the sum of the distances is minimum along the perpendicular bisector of the line segment $$AB$$; the actual minimum is at the point of intersection of the bisector and $$y=x+3$$, but since you have an additional constraint, you find the intersection of the bisector with $$y=x+2$$, call it $$C$$.

Note that if you drop perpendiculars to $$x$$ and $$y$$ axes respectively from $$A$$ and $$B$$, they intersect at $$A'=(0,2)$$ and $$B'=(2,4)$$. You'll notice that their mid point is $$C$$. Then $$C$$ turns out to be $$(1,3)$$. That gives us that $$\alpha=20$$.

• Ah, I solved through this method. See the comments on @Robert . Thanks for the help. Dec 2, 2018 at 16:48

Your method is also really good. Consider the two points $$A(-1,2)$$ and $$B(2,5)$$. We need to find the point $$(x,x+2)$$ on the line $$y=x+2$$ for which the sum $$AD+BD$$ is minimum (see the graph below).

$$\hspace{3cm}$$

You reflect the point $$A$$ over the line $$y=x+2$$ to find the point $$C(0,1)$$. The line through $$C$$ and $$B$$ is $$y=2x+1$$. The two lines intersect at $$D(1,3)$$, which is the solution of the given problem. Hence, when $$x=1$$, the sum will be minimum $$\sqrt{20}$$.

• Thanks for another method (2nd para). Dec 2, 2018 at 16:59
• You are welcome. It is actually the method you started and asked suggestion to continue. Good luck. Dec 2, 2018 at 17:06

By letting $$x=t+1$$ we get an even function $$\sqrt{2t^2+6t+5}+\sqrt{2t^2-6t+5}.$$ Now we show that the minimum value is attained at $$t=0$$: we have to verify $$\sqrt{2t^2+6t+5}+\sqrt{2t^2-6t+5}\geq \sqrt{20}$$ or, after squaring, $$4t^2+10+2\sqrt{4t^4-16t^2+25}\geq 20$$ that is $$\sqrt{(5-2t^2)^2+4t^2}\geq 5-2t^2$$ which trivially holds.

• $\alpha$=20....? Dec 2, 2018 at 16:33
• Yes, now try to prove it. Dec 2, 2018 at 16:36
• $\alpha=20$ is correct by I'd prefer an algebraic approach. Dec 2, 2018 at 16:45
• The symmetry in $t$ gives you the minimum candidate. The proof is up to you (algebraic or geometric). Dec 2, 2018 at 16:51
• @jayant98 See my (algebraic) conclusion. Dec 2, 2018 at 16:56

Hint.

Calling $$f(x) = \sqrt{x^2+(x+1)^2}$$ we seek for

$$\min_x f(x) + f(x-3)$$

If we write this as $$\sqrt{x^2+(x+1)^2}+\sqrt{(x-2)^2+(x-3)^2}$$ then we are searching for a point $$T$$ on line $$y=x$$ for which $$TA+TB$$ takes minumim where $$A(0,-1)$$ and $$B(3,2)$$.

Now this is well know problem froma ancient greek. Reflect $$A$$ acros this line and get $$A'(-1,0)$$. By triangle inequality we can see that $$TA+TB\geq A'B$$ and that minimum is achieved at intersection of lines $$y=x$$ and line $$A'B$$ which is $$y = {x +1 \over 2}$$.

So $$\alpha = A'B^2 = 20$$.