Maximize
f = 30x + 27y
Subject To:
30x + 10y ≤ 180
20x + 20y ≤ 200
x≥0, y≥0
a.) Graph the solution of the constraint system
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≥0, y≥0 tells us.
We will also find the point where the slanted lines intersect
by solving this system:
Use substitution or elimination to get x=4, y=6, so they intersect
at the point (x,y) = (4,6)
b.) Find the corners of the resulting feasible region
Since the first two inequalities have ≤, 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:
c.) Evaluate the objective function at each corner
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
d.) Indicate the corner which provides the Maximum value for the objective function.
The maximum value is 282 which is reached at the corner point (4,6).
Edwin
