What is the point in the feasible region maximizes the objection function?

First, find the vertices of the shaded region. You can do this my graphing or by solving a system of equations for each pair of functions. I am going to find them by graphing.

x ≥ 0 and y ≥ 0 place the shaded region in the first quadrant with a vertex at (0, 0)

now graph y ≤ 3.  a vertex is at (0, 3)

now graph y < -2x + 5. a vertex is at (1, 3). the other vertex is the x-intercept (when y = 0).

                  0 = -2x + 5

                  -5 = -2x

                        So, the vertex is  

Next, input the coordinates of the vertices into the objective function.

C = -6x + 5y

(0, 0): C = -6(0) + 5(0)

             =    0    +   0

             =          0

(0, 3): C = -6(0) + 5(3)

             =    0    +   15

             =          15

(1, 3): C = -6(1) + 5(3)

             =  -6  +  15

             =       9

: C = -6(\frac{5}{2}[/tex]) + 5(2)

             =  -15  +  10

             =       -5

The maximum is C = 15 which occurs at vertex (0, 3)

A

the answer is 1/4

step-by-step explanation: