A linear programming feasible region is determined by
6x + 3y < 72
5x + 15y < 185
x > 0, y > 0
Which of the following objective functions has
its maximum value at the intersection of
6x + 3y = 72 and
5x + 15y = 185
A. Z = 10x + 10y
B. Z = 5x + 2y
C. Z = 2x + 8y
Graph the four boundary lines:
6x + 3y = 72 (6x + 3y < 72 will
be the region on or below this line)
5x + 15y = 185 (5x + 15y < 185
will be the region on or below this
line.
x = 0 (this is the y-axis, x > 0
will be the region to the right of the
y-axis.
y = 0 (this is the x-axis, y > 0
will be the region above the x-axis.
You must solve this system
6x + 3y = 72
5x + 15y = 185
To find out where the lines intersect:
I assume you can do this, the intersection
is the point (x,y) = (7,10)
A. Z = 10x + 10y
B. Z = 5x + 2y
C. Z = 2x + 8y
Now we'll have to try each of these out to find out if
the maximum value of Z occurs at the point (7,10)
corner point | x | y | z = 10x + 10y |
--------------------------------------------------
(0,37/3) | 0 |37/3 | 10(0)+10(37/3)= 370/3 = 123.33...
(0,0) | 0 | 0 | 10(0)+10(0) = 0
(12,0) | 12 | 0 | 10(12)+10(0) = 120
(7,10) | 7 | 10 | 10(7)+10(10) = 170
This is a correct one, since 170 is the largest o
value and that occurs at (7,10). Let's check the other
two:
corner point | x | y | z = 5x + 2y |
--------------------------------------------------
(0,37/3) | 0 |37/3 | 5(0)+2(37/3)= 74/3 = 24.66...
(0,0) | 0 | 0 | 5(0)+2(0) = 0
(12,0) | 12 | 0 | 5(12)+2(0) = 60
(7,10) | 7 | 10 | 5(7)+2(10) = 55
No, that's not an answer, because 55, which occurs at
(7,10), is not the largest value. Let's try the third one:
corner point | x | y | z = 2x + 8y |
--------------------------------------------------
(0,37/3) | 0 |37/3 | 2(0)+8(37/3)= 296/3 = 98.66...
(0,0) | 0 | 0 | 2(0)+8(0) = 0
(12,0) | 12 | 0 | 2(12)+8(0) = 24
(7,10) | 7 | 10 | 2(7)+9(10) = 104
This is a correct answer, too since 104 is the largest of the
four values and that occurs at (7,10).
So the answer is A and C
Edwin
