Find the minimum value of $|x+1|+2|x-5|+|2x-7|+|\frac{x-11}{2}|$ 
Find the minimum value of $|x+1|+2|x-5|+|2x-7|+|\frac{x-11}{2}|$.

I have no idea how to approach this question.  However, I managed to solve it using a rather childish approach.   I change this equation by multiplying $2$, getting $2|x+1|+4|x-5|+4|x-3.5|+|x-11|$.  Now, I have a number line, and I mark out $-1$, $3.5$, $5$ and $11$.  Here, I shall use this analogy which may seem childish.  There are $2$ houses at $-1$, $4$ houses at $3.5$, $4$ houses at $5$, and $1$ house at $11$.  
Here, I introduce a train station.  We need to place it at an optimum place to minimise the distance from every house to it.  If it is between $3.5$ and $5$, observe that it needs to move left to please more people(easier to board train).  If it is placed between $-1$ and $3.5$, it needs to move right to please more people.  Hence, the optimum place to put the train station would be at $3.5$, hence I have $x=3.5$, which is correct.  You guys can draw a diagram to understand better.  In addition, I noted that this seemed rather similar to some programming problems.  
Back to the question, is there a much more elegant way to this problem?

Case 1:$x\le -1\\f(x)=-x-1+10-2x+7-2x+5.5-0.5x=21.5-5.5x$
Case 2:$-1\le x\le 3.5\\f(x)=x+1+10-2x+7-2x+5.5-0.5x=23.5-3.5x$
Case 3:$3.5\le x\le 5\\f(x)=x+1+10-2x-7+2x+5.5-0.5x=9.5+0.5x$
Case 4:$5\le x\le 11\\f(x)=x+1-10+2x-7+2x+5.5-0.5x=-10.5+4.5x$
Case 5:$11\le x\\f(x)=x+1-10+2x-7+2x+0.5x-5.5=-21.5+5.5x$
so we have: $$\min_{x\in\Bbb R}f(x)=\min_{x\le-1}f(x)+\min_{-1\le x\le 3.5}f(x)+\min_{3.5\le x\le 5}f(x)+\min_{5\le x\le 11}f(x)+\min_{x\ge 11}f(x)=\min\lbrace{27,11.25,11.25,12,39}\rbrace=11.25$$
here is a sketch:

