Answer to Question #149826 in Statistics and Probability for Wesam

Question

Let x, y, and z be nonnegative real numbers with x + y + z = 120. Compute the largest possible value of the median of the three numbers 2x + y, 2y + z, and 2z + x.

Answer 1

We express $x$ and receive: $240-y-2z$, $2y+z$, $120+z-y$ . We remind that a median is a number that splits the set into the higher half and lower half. From equation it is clean that any variable (x,y or z) can not take values higher than 120. In case we set: $z=0$, $y=80$ we receive: $160,160,40$. The median will be equal to 160. The total sum (for any choice if values) is equal to $360$. From the latter we conclude that the median can not be higher than 180. Consider the sum of the first two numbers: $240+y-z$. The last one is $120+z-y$ . In case $y<z$, in order to achieve the median equal to 160, the follwoing equalities have to be satisfied: $y+2z<80$, $2y+z>160$ From the latter we receive: $-y-2z>-80$. In case we add both equalities, one receives: $y-z>80$ . It means that $y>80$, but then the first number will be less than 160. Thus, the answer is 160.