Minimization of function with two parameters.

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Let this minimization, $$ \min_{x\in[-2y,4y]}((1-(y+\frac{1}{2}x))^++(5/4-(y+\frac{1}{4}x))^++(7/4-(y-\frac{1}{4}x))^+)\ \ \ \ \ \ \ \ (*) $$ with $y\geq 0$, $x\in\mathbb R$ and the notation $(.)^+$ means : $(z)^+=\max(0,z)$.

Is there a solution of $(*)$ as in this example :

$$ \min_{x\in[-y/2,y]} [(-(y+2x))^++(-(y+x))^++(1-(y-x))^+]=(1-3/2y)^+ $$ Where the minimum is attained at $x = -y/2$.

I have tried by the mean of differentiation without success.

Is $x=-2y$ the solution of $(*)$ ? or it depends to $y$ ?

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This problem can be arranged as a LP program

$$ \min_{x,y,z_1,z_2,z_3} z_1+z_2+z_3 \;\;\mbox{s. t.} $$ $$ z_1=1-y-\frac x2\\ z_2=\frac 54-y-\frac x4\\ z_3=\frac 74-y+\frac x4\\ z_1 \ge 0\\ z_2 \ge 0\\ z_3 \ge 0\\ -2y \le x\\ x\le 4y\\ y \ge 0 $$

with one solution at

$$ x = -1, y = \frac 32, z_1 = 0, z_2 = 0, z_3 = 0 $$

Attached a plot showing in black the level surfaces for $f(x,y) = \max( 0,z_1)+\max(0,z_2)+\max(0,z_3)\;$ in red the lines $z_1=0, z_2=0, z_3=0$

enter image description here