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