Question 971810
Maximize
f = 30x + 27y
Subject To:
30x + 10y ≤ 180
20x + 20y ≤ 200
x≥0, y≥0
a.) Graph the solution of the constraint system 
<pre>
The boundary lines' equations:
30x + 10y = 180 
20x + 20y = 200
        x = 0  (the equation of the y-axis)
        y = 0  (the equation of the x-axis)

These simplify to
3x + y = 18 which has intercepts (0,18) and (6,0)
 x + y = 10 which has intercepts (0,10) and (10,0)

We only need to draw the 1st quadrant since neither variable
is negative, which is what x&#8805;0, y&#8805;0 tells us.

We will also find the point where the slanted lines intersect
by solving this system:

{{{system(3x+y=18,x+y=10)}}}

Use substitution or elimination to get x=4, y=6, so they intersect 
at the point (x,y) = (4,6)

{{{drawing(2800/11,400,-2,12,-2,20,line(0,0,20,0),line(0,0,0,25),
locate(.2,18.5,"(0,18)"),locate(4.2,6.6,"(4,6)"),
locate(.2,10.5,"(0,10)"),locate(9.8,1,"(10,0)"),locate(6,1,"(6,0)"),
locate(.3,1,"(0,0)"),
line(0,18,6,0), line(10,0,0,10) )}}}
</pre>
b.) Find the corners of the resulting feasible region
<pre>
Since the first two inequalities have &#8804;, the feasible region is the region
below the two slanted lines, to the right of the y-axis and above the x-axis.
Let's eliminate all of the graph except for the feasible region:

{{{drawing(2800/11,400,-2,12,-2,20,line(0,0,20,0),line(0,0,0,25),locate(4.2,6.6,"(4,6)"),locate(.3,1,"(0,0)"),
locate(.2,10.5,"(0,10)"),locate(6,1,"(6,0)"),

line(4,6,6,0), line(4,6,0,10) )}}}
</pre>
c.) Evaluate the objective function at each corner 
<pre>
Corner |  
point  |f(x) = 30(x) + 27(y)  =    VALUE      
-----------------------------------------------
(0,0)  |       30(0) + 27(0)  =   0 +   0 =   0   <---minimum value = 0
(0,10) |       30(0) + 27(10) =   0 + 270 = 270
(4,6)  |       30(4) + 27(6)  = 120 + 162 = 282   <---maximum value = 282
(6,0)  |       30(6) + 27(0)  = 180 +   0 = 180
</pre>
d.) Indicate the corner which provides the Maximum value for the objective function.
<pre>
The maximum value is 282 which is reached at the corner point (4,6).

Edwin</pre>