Find the minimum value of |x−y|+√(x+2)2+(y−4)4
Problem
Source:
Tags: algebra unsolved, algebra
21.06.2011 18:31
shohvanilu wrote: Find the minimum value of |x−y|+√(x+2)2+(y−4)4 You should change your source of olympiad training problems (or stop just inventing crazy problems) Replacing x→x−2 and y→a+4, we can define f(x)=|x−a−6|+√x2+a4 If x≥a+6, we get f′(x)=x+√x2+a4√x2+a4≥0 and so minimum is reached when x=a+6 If x≤a+6, we get f′(x)=x−√x2+a4√x2+a4≤0 and so minimum is reached when x=a+6 f(a+6)=√(a+6)2+a4 and we are looking for a such that this value is minimum So we are looking for the minimum of g(x)=x4+(x+6)2=x4+x2+12x+36 g′(x)=4x3+2x+12 g"(x)=12x^2+2>0 So g(x) is decreasing on (-\infty,r] and increasing on [r,+\infty) and the requested value is \sqrt{g(r)} where r is the unique real root of x^3+\frac 12x+3=0 This is a standard basic cubic in Cardano's form and we immediately get r=\sqrt[3]{\frac{\sqrt{2922}-54}{36}} -\sqrt[3]{\frac{\sqrt{2922}+54}{36}} Hence the answer : Minimum is \sqrt{r^4+r^2+12r+36} where r=\sqrt[3]{\frac{\sqrt{2922}-54}{36}} -\sqrt[3]{\frac{\sqrt{2922}+54}{36}} Numeric application gives : r\sim -1.32695628567897... \min_{x,y\in\mathbb R}(|x-y|+\sqrt{(x+2)^2+(y-4)^4} \boxed{\sim 4.99377611842389...}
22.06.2011 00:40
I hope he meant |x-y|+\sqrt{(x+2)^2+(y-4)^2} (the power 4 was just a typo); then at least it has a nice geometric meaning and a nice, tame solution.
22.06.2011 08:33
mavropnevma wrote: I hope he meant |x-y|+\sqrt{(x+2)^2+(y-4)^2} (the power 4 was just a typo); then at least it has a nice geometric meaning and a nice, tame solution. Maybe it was |x-y|+\sqrt{(x+2)^0+(y-4)^0}, which seems easy too.
05.01.2013 10:42
is there a solution for |x-y|+\sqrt{(x+2)^2+(y-4)^2}
06.01.2013 11:42
You can fix |x-y|=r>0. Then for each r, let f(r) be the minimum distance (in r) of any line in the curve |x-y|=r to the point (-2,4). Then, minimize r+f(r).
13.06.2014 06:58
Mavropnevma you right it's original version of this problem.
14.03.2016 02:38
Sardor wrote: Mavropnevma you right it's original version of this problem. |x-y|+\sqrt{(x+2)^2+(y-4)^2}\ge|x-y|+\frac{\sqrt{2}}{2}|(x+2)-(y-4)| =|x-y|+|3\sqrt{2}+\frac{\sqrt{2}}{2}(x-y)|\ge |x-y|+3\sqrt{2}+\frac{\sqrt{2}}{2}(x-y) \ge (1-\frac{\sqrt{2}}{2})|x-y|+3\sqrt{2}\ge 3\sqrt{2}
25.03.2017 05:34
CAn we substitute x by cost-2 and y by sint +4?