The objective function is given by `f(x,y)=6x+3y` subject to the following constraints:

`2x+2y>=4` (1)

`-x+5y<=10` (2)

`3x-3y<=6` (3)

`x>=0,y>=0`

The last constraints mean we are only interested in the first quadrant.

We graph the constraints to find the feasible region. The maximum and minimum of teh objective function occur...

## Unlock

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

The objective function is given by `f(x,y)=6x+3y` subject to the following constraints:

`2x+2y>=4` (1)

`-x+5y<=10` (2)

`3x-3y<=6` (3)

`x>=0,y>=0`

The last constraints mean we are only interested in the first quadrant.

We graph the constraints to find the feasible region. The maximum and minimum of teh objective function occur at the vertices of the feasible region.

Note that I cannot shade the feasible region:

The feasible region is the interior of the triangle. Since the feasible region is bounded, it has a maximum and a minimum.

The vertices are (0,2),(2,0),(5,3)

`f(0,2)=6(0)+3(2)=6` This is the minimum

`f(2,0)=6(2)+3(0)=12`

`f(5,3)=6(5)+3(3)=39` This is the maximum