Question 1207825
<pre>
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 
<font color="red">3x+2y≤12</font>  
<font color="blue">2x+y≤7</font>  
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, <font color="red">3x+2y≤12</font>  and <font color="blue">2x+y≤7</font> . Use the origin 
(x,y) = (0,0) as a test point to find which side of each line the
shading will appear on.

Testing <font color="red">3x+2y≤12: 3(0)+2(0)≤12</font> is true so we shade the side of that
line which the origin is on, which is underneath the line:

Testing <font color="blue">2x+y≤7: 2(0)+(0)≤7</font> 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 
<font color="red">3x+2y=12</font> 
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 
<font color="blue">2x+y=7</font> 
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:

{{{system(blue(3x+2y=12), red(2x+y=7))}}}

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

{{{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,-.9,7,12,13,(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)))}}}

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</pre>