Let there be x cakes of first kind and y cakes of second kind. Therefore, x≥0 and y≥0The given information can be complied in a table as follows. ∴200x+100y≤5000 ⇒2x+y≤50 25x+50y≤1000 ⇒x+2y≤40 Total numbers of cakes, z, that can be made are, z = x + y The mathematical formulation of the given problem isMaximize z = x + y ........ (1)subject to the constraints,2x+y≤50 ........ (2)x+2y≤40 ........ (3)x,y≥0 ........ (4)The feasible region determined by the system of constraints is as follows.The corner points are A(25,0), B(20,10), O(0,0), and C(0,20). The values of Z at these corner points are as follows.Thus, the maximum numbers of cakes that can be made are 30 ( 20 of one kind and 10 of the other kind).