Idon't know how to do this! given the system of constraints, name all vertices of the feasible region. then find maximum value of the given objective function. constraints {x≥0, y≥0, y≤1/3x+3, 5≥y+xobjective function: c = 6x-4y show

Constraints (in slope-intercept form)
x≥0,
y≥0,
y≤1/3x+3,
y≤ 5 - x

The vertices are the points of intersection between the constraints, or the outer bounds of the area that agrees with the constraints.
We know that x≥0 and y≥0, so there is one vertex at (0,0)
We find the other vertex on the y-axis, plug in 0 for x in the function:
y ≤ 1/3x+3
y ≤1/3(0)+3
y = 3. 
There is another vertex at (0,3)
Find where the 2 inequalities intersect by setting them equal to each other
(1/3x+3) = 5-x          Simplify Simplify Simplify
x = 3/2
Plugging in 3/2 into y = 5-x: 10/2 - 3/2 = 7/2
y=7/2
There is another vertex at (3/2, 7/2)
 
There is a final vertex where the line y=5-x crosses the x axis:
0 = 5 -x , x = 5
The final vertex is at point (5, 0)
Therefore, the vertices are:
(0,0), (0,3), (3/2, 7/2), (5, 0)
We want to maximize C = 6x - 4y.
Of all the vertices, we want the one with the largest x and smallest y. We might have to plug in a few to see which gives the greatest C value, but in this case, it's not necessary. 
The point (5,0) has the largest x value of all vertices and lowest y value.
Maximum of the function:
C = 6(5) - 4(0)
C = 30

5(2a-5+b). xbox xboxxbx xb

Main idea on left, details on right