Let X and Y be the numbers of model A sets and model B sets, respectively.


Then the constrains are


30 <= X <= 60,
10 <= Y <= 30,

X + Y <= 60.


The profit function is C(X,Y) = 40*X + 60*Y  dollars to maximize.


The feasibility area is shown in the Figure below






Plots  X = 30 (red),  X = 60 (green),  Y = 10 (blue),  Y = 30 (magenta) and  X + Y = 60.



The feasibility domain is the triangle area bounded by the red, blue and sloped lines.


The points to check the cost function are

    P1 = (30,10)    (right angle vetex)

    P2 = (30,30)    (upper vertex of the triangle)

    P3 = (50,10)    (most right vertex of the triangle).


The values of the profit function are


    at P1:  P(30,10) = 30*40 + 10*60 = 1800;

    at P2:  P(30,30) = 30*40 + 30*60 = 3000;

    at P3:  P(50,10) = 50*40 + 10*60 = 2600.


You are looking for the maximum - hence, your solution is at P2.


Answer.  30 model A sets and 30 model B sets is the solution which provides the maximal profit of $3000 per day.