What is the perimeter of the boundary of the region consisting of all points which can be expressed as $(2u-3w, v+4w)$ with $0\le u\le1$, $0\le v\le1,$ and $0\le w\le1$?
$\textbf{(A) } 10\sqrt{3} \qquad \textbf{(B) } 13 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 16$
Notice that this we are given a parametric form of the region, and $w$ is used in both $x$ and $y$. We first fix $u$ and $v$ to $0$, and graph $(-3w,4w)$ from $0\le w\le1$. When $w$ is $0$, we have the point $(0,0)$, and when $w$ is $1$, we have the point $(-3,4)$. We see that since this is a directly proportional function, we can just connect the dots like this: 
 [asy] 	import graph; 	Label f; size(5cm); 	unitsize(0.7cm);  	xaxis(-5,5,Ticks(f, 5.0, 1.0)); 	yaxis(-5,5,Ticks(f, 5.0, 1.0));  	draw((0,0)--(-3,4)); 	[/asy]
	Now, when we vary $u$ from $0$ to $2$, this line is translated to the right $2$ units: 
 [asy] 	import graph; 	Label f;  	unitsize(0.7cm); size(5cm); 	xaxis(-5,5,Ticks(f, 5.0, 1.0)); 	yaxis(-5,5,Ticks(f, 5.0, 1.0));  	draw((0,0)--(-3,4)); 	draw((2,0)--(-1,4)); 	[/asy]
	We know that any points in the region between the line (or rather segment) and its translation satisfy $w$ and $u$, so we shade in the region: 
 [asy] 	import graph; 	Label f;  	unitsize(0.7cm); size(5cm); 	xaxis(-5,5,Ticks(f, 5.0, 1.0)); 	yaxis(-5,5,Ticks(f, 5.0, 1.0));  	draw((0,0)--(-3,4)); 	draw((2,0)--(-1,4));  	filldraw((0,0)--(-3,4)--(-1,4)--(2,0)--cycle, gray); 	[/asy]
	We can also shift this quadrilateral one unit up, because of $v$. Thus, this is our figure: 
 [asy] 	import graph; 	Label f;  	unitsize(0.7cm); size(5cm); 	xaxis(-5,5,Ticks(f, 5.0, 1.0)); 	yaxis(-5,5,Ticks(f, 5.0, 1.0));  	draw((0,0)--(-3,4)); 	draw((2,0)--(-1,4));  	filldraw((0,0)--(-3,4)--(-1,4)--(2,0)--cycle, gray); 	filldraw((0,1)--(-3,5)--(-1,5)--(2,1)--cycle, gray);  draw((0,0)--(0,1),black+dashed); draw((2,0)--(2,1),black+dashed); draw((-3,4)--(-3,5),black+dashed); 	[/asy]
 [asy] 	import graph; 	Label f;  	unitsize(0.7cm); size(5cm); 	xaxis(-5,5,Ticks(f, 5.0, 1.0)); 	yaxis(-5,5,Ticks(f, 5.0, 1.0));  	draw((0,0)--(-3,4)); 	draw((1,0)--(-2,4));  	filldraw((0,0)--(2,0)--(2,1)--(-1,5)--(-3,5)--(-3,4)--cycle, gray); 	[/asy]
	The length of the boundary is simply $1+2+5+1+2+5$ ($5$ can be obtained by Pythagorean theorem, since we have side lengths $3$ and $4$.). This equals $\boxed{\textbf{(E) }16.}$
~Technodoggo ~ESAOPS