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
