The maximum value of z = 5x + 4y subject to 

3x + y < 24
6x + 4y < 66
x > 0, y > 0 is 

A. 96
B. 66
C. 56
D. 40

Graph the boundary lines:

1.     3x + y = 24  (3x + y < 24 will 
be the region on or below this line)



2.     6x + 4y = 66  (6x + 4y < 66 will 
be the region on or below this line)



3.     x = 0  (x > 0 will be the 
region on or to the right of this line, 
which is just the y-axis.

  

4.     y = 0  (y > 0 will be the 
region on or above this line, which is 
just the x-axis.

 

You can shade the common region.  I can't shade 
on here so I will just erase all the parts of 
the lines that I don't need:



Now we will find all four corner points.

The top point is found by solving the system

6x + 4y = 66
x = 0

That has the solution (0,16.5)

The bottom left point is obviously the 
origin but is found by solving the system

x = 0
y = 0

That has solution (0,0)

The bottom right point is found by solving
the system

3x + y = 24
y = 0 

That has solution (8,0)

The point in the middle is found by solving 
the system

3x + y = 24
6x + 4y = 66

That has solution (5,9)



Now both the maximum and the minimum values of
the objective function 

z = 5x + 4y

will occur at corner points. So we make this table:

corner point |  x  |  y  |    z = 5x + 4y   |
---------------------------------------------
   (0,16.5)  |  0  |16.5 |  5(0)+4(16.5)=66   
    (0,0)    |  0  |  0  |  5(0)+4(0) =   0
    (8,0)    |  8  |  0  |  5(8)+4(0) =  40
    (5,9)    |  5  |  9  |  5(5)+4(9) =  61

So we find that the maximum value of the
objective function z is 66 when x=0 and y=16.5
(and the minimum value is 0 when x=0 and y=0).

But you wanted the maximum value so it's

z = 66 when x=0 and y = 16.5, which is
choice B.

Edwin