What she means is that you should have ≤, not <, and ≥, not >.
Otherwise your inequalities will not include the boundary lines,
and will have no maximum (or minimum) solutions.

Maximize z=x+5y 
subject to 
3x+2y≤12  
2x+y≤7  
x≥0, y≥0

Here is the first quadrant graphs of the lines whose equations
are like the constraint inequalities with equal signs instead of
the inequality symbols, 3x+2y≤12  and 2x+y≤7 . Use the origin 
(x,y) = (0,0) as a test point to find which side of each line the
shading will appear on.

Testing 3x+2y≤12: 3(0)+2(0)≤12 is true so we shade the side of that
line which the origin is on, which is underneath the line:

Testing 2x+y≤7: 2(0)+(0)≤7 is true so we shade the side of that
line which the origin is on, which is also underneath the line.
so we shade under BOTH lines.

The feasible region is the shaded area.

We find all the corner points. by finding the x and y intercepts of both
lines and their point of intersection.  

The y-intercept of 
3x+2y=12 
is found by substituting x=0 and solving for y,
we get y-intercept for the red line is (0,6)

The x-intercept of 
2x+y=7 
is found by substituting y=0 and solving for x,
we get x-intercept for the blue line is (3.5,0)  

We get the corner point which is the intersection of
the two lines:



Solve that by substitution or elimination and get (2,3)



Now we find the value of the objective function

z = x+5y at each of the 4 corner points:

At corner point (x,y) = (0,0),  z = x+5y = 0+5(0) = 0.

At corner point (x,y) = (0,6),  z = x+5y = 0+5(6) = 30.

At corner point (x,y) = (2,3),  z = x+5y = 2+5(3) = 17.

At corner point (x,y) = (3.5,0),  z = x+5y = 3.5+5(0) = 3.5.

So z has a maximum value of 30 when x=0 and y=6.

Edwin

What Greenestamps means is this:

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

 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:



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