Question 1207825
<pre>
What Greenestamps means is this:

The objective function z=x+5y can be written in slope intercept form y=mx+b

{{{y=expr(-1/5)x+z}}} where -1/5 is the slope, and z is the y intercept.  
Think of all the parallel lines that have slope -1/5. 
I'll draw a bunch of them across my graph:

{{{drawing(27600/79,400,-.9,6,-.9,7,
graph(27600/79,400,-.9,6,-.9,7,((12-3x)/2)*sqrt(x)/sqrt(x)*sqrt(4-x)/sqrt(4-x)),

graph(27600/79,400,-.9,6,-.9,7,(-1/5)x-1),
graph(27600/79,400,-.9,6,-.9,7,(-1/5)x+0),
graph(27600/79,400,-.9,6,-.9,7,(-1/5)x+1),
graph(27600/79,400,-.9,6,-.9,7,(-1/5)x+2),
graph(27600/79,400,-.9,6,-.9,7,(-1/5)x+3),
graph(27600/79,400,-.9,6,-.9,7,(-1/5)x+4),
graph(27600/79,400,-.9,6,-.9,7,(-1/5)x+5),
graph(27600/79,400,-.9,6,-.9,7,(-1/5)x+6),
graph(27600/79,400,-.9,6,-.9,7,(-1/5)x+7),
graph(27600/79,400,-.9,6,-.9,7,(-1/5)x+8),
graph(27600/79,400,-.9,6,-.9,7,(-1/5)x+9),
graph(27600/79,400,-.9,6,-.9,7,(-1/5)x+10),
graph(27600/79,400,-.9,6,-.9,7,(-1/5)x+11),



graph(27600/79,400,-.9,6,-.9,7,12,13,(7-2x)*( sqrt(3.5-x)/sqrt(3.5-x) )),

locate(2.1,3.2,"(2,3)"), locate(-1,6.3,"(0,6)"),locate(-.7,0,"(0,0)"),
locate(3,-.3,"(3.5,0)"),
  
green(

line(0,0,0,6),line(0.05,0,0.05,5.925),line(0.1,0,0.1,5.85),line(0.1,0,0.1,5.85),
line(0.15,0,0.15,5.775),line(0.2,0,0.2,5.7),line(0.25,0,0.25,5.625),line(0.25,0,0.25,5.625),
line(0.3,0,0.3,5.55),line(0.35,0,0.35,5.475),line(0.4,0,0.4,5.4),line(0.4,0,0.4,5.4),
line(0.45,0,0.45,5.325),line(0.5,0,0.5,5.25),line(0.55,0,0.55,5.175),line(0.55,0,0.55,5.175),
line(0.6,0,0.6,5.1),line(0.65,0,0.65,5.025),line(0.7,0,0.7,4.95),line(0.7,0,0.7,4.95),
line(0.75,0,0.75,4.875),line(0.8,0,0.8,4.8),line(0.85,0,0.85,4.725),line(0.85,0,0.85,4.725),
line(0.9,0,0.9,4.65),line(0.95,0,0.95,4.575),line(1.0,0,1.0,4.5),line(1.0,0,1.0,4.5),
line(1.05,0,1.05,4.425),line(1.1,0,1.1,4.35),line(1.15,0,1.15,4.275),line(1.15,0,1.15,4.275),
line(1.2,0,1.2,4.2),line(1.25,0,1.25,4.125),line(1.3,0,1.3,4.05),line(1.3,0,1.3,4.05),
line(1.35,0,1.35,3.975),line(1.4,0,1.4,3.9),line(1.45,0,1.45,3.825),line(1.45,0,1.45,3.825),
line(1.5,0,1.5,3.75),line(1.55,0,1.55,3.675),line(1.6,0,1.6,3.6),line(1.6,0,1.6,3.6),
line(1.65,0,1.65,3.525),line(1.7,0,1.7,3.45),line(1.75,0,1.75,3.375),line(1.75,0,1.75,3.375),
line(1.8,0,1.8,3.3),line(1.85,0,1.85,3.225),line(1.9,0,1.9,3.15),line(1.9,0,1.9,3.15),
line(1.95,0,1.95,3.075),

line(2,0,2,3),line(2.05,0,2.05,2.9),line(2.1,0,2.1,2.8),line(2.1,0,2.1,2.8),
line(2.15,0,2.15,2.7),line(2.2,0,2.2,2.6),line(2.25,0,2.25,2.5),line(2.25,0,2.25,2.5),
line(2.3,0,2.3,2.4),line(2.35,0,2.35,2.3),line(2.4,0,2.4,2.2),line(2.4,0,2.4,2.2),
line(2.45,0,2.45,2.1),line(2.5,0,2.5,2.0),line(2.55,0,2.55,1.9),line(2.55,0,2.55,1.9),
line(2.6,0,2.6,1.8),line(2.65,0,2.65,1.7),line(2.7,0,2.7,1.6),line(2.7,0,2.7,1.6),
line(2.75,0,2.75,1.5),line(2.8,0,2.8,1.4),line(2.85,0,2.85,1.3),line(2.85,0,2.85,1.3),
line(2.9,0,2.9,1.2),line(2.95,0,2.95,1.1),line(3.0,0,3.0,1.0),line(3.0,0,3.0,1.0),
line(3.05,0,3.05,0.9),line(3.1,0,3.1,0.8),line(3.15,0,3.15,0.7),line(3.15,0,3.15,0.7),
line(3.2,0,3.2,0.6),line(3.25,0,3.25,0.5),line(3.3,0,3.3,0.4),line(3.3,0,3.3,0.4),
line(3.35,0,3.35,0.3),line(3.4,0,3.4,0.2),line(3.45,0,3.45,0.1),line(3.45,0,3.45,0.1),
line(3.5,0,3.5,0.000000000000010658141)))}}}

You see that the only two of those parallel lines that intersect the feasible
region in only one corner points are the ones with y-intercept 6 and 0.
So the maximum value of z is 6 (when x=0, y=6) and the minimum value of z 
is 0 (when x=0, y=0).  Many times you can just visualize mentally how steep,
or how non-steep, the objective function slants, whether it slants upward
or downward and determine by inspection which are the points at which the
maximum and minimum values of z occur without substituting the corner points in
the objective function.   

But be aware that it won't always be the top and bottom ones like it is in this
particular example.  The shapes of feasible regions differ greatly. 

Edwin</pre>