Let X be the number of $640-refrigerators 
and Y be the number of $780-refrigerators.


Then the Revenue function is  R(X,Y) = 640*X + 780*Y.    (1)


The constraints are 

X + Y <= 250,     (2)  

X >= 80           (3)

Y >= 120.         (4)


The feasible domain is the triangle in the first quadrant coordinate (X,Y)-plane restricted by the straight lines (2), (3) band (4).

The vertices of this triangle are   (MAKE YOUR SKETCH !)

    P1 = (80, 120),
    P2 = (80, 170),    ( <<<---===   notice that 80  + 170 = 250 ! )
    P3 = (130,120)     ( <<<---===   notice that 130 + 120 = 250 ! )


You need to maximize the function (1) over this triangle.


Apply the linear programming method.
The method says: 

    To find the maximum, calculate the function F(X,Y) at the vertices of the polygon (at the vertices of the triangle, in this case !)

    and select that vertex where the function is maximal.


So,  at  P1:  F( 80,120) =  80*640 + 120*780 = 144800;

     at  P2:  F( 80,170) =  80*640 + 170*780 = 183800;

     at  P3:  F(130,120) = 130*640 + 120*780 = 176800.


As you see, the function F(X,Y) is maximal at P2.

Hence,  X = 80  and  Y = 170 is the optimal solution.


Answer.  80 $640 refrigerators and 170 $780 refrigerators provide the maximal revenue of $183800.