Let X be the number of standard chairs produced each day, and let y be the number of plush chairs produced.


Then you have these restrictions

2X + 3y <= 240              (1)   (hours to construct and finish)   and
 X + 3Y <= 150              (2)   (hours to upholstering)

The revenue function is 

R(X,Y) = 89*X + 135.5*Y     (3)

Two other restrictions are

X >= 0,  Y >= 0.


Restricting straight lines are shown in the figure below.






Plot 2X + 3Y = 240  (red)  and  X + 3Y = 150 (green)



The feasibility area is the quadrilateral in the first quadrant restricted by the axes and the red and green straight  lines.


The task is to find the maximum value of the Revenue function (3) over this quadrilateral.


The method of linear programming says:

    1) Take the vertices of this quadrilateral

        (x1,Y1) = (0,50)    (green line Y-intercept)
        (x2,Y2) = (120,0)   (red line X-intercept)
        (x3,Y3) = (90,20)   (intersection point of the straight lines Y =  and Y =  )
        (x4,Y4) = (0,0)     (the origin of the coordinate system)

    2) Calculate the objective function at these points

        R(X1,Y1) = 89*0   + 135.5*50 = 6775;

        R(X2,Y2) = 89*120 + 135.5*0  = 10680;

        R(X3,Y3) = 89*90  + 135.5*20 = 10720.

        R(X4,Y4) = 89*0   + 135.5*0  = 0.


    3) Then select one of these point where the objective function is maximal - In our case this point is (X3,Y3) = (90,20)


    4) This point gives your optimal solution X = 90 of standard chairs and Y = 20 of plush chairs.


If they follow this optimal solution, their daily revenue will be MAXIMAL, $10720.