A farmer has 100 acres of land on which she can grow wheat and corn. She has a total of 1200 man-hours of labor available. It takes 6 man-hours to cultivate an acre of wheat and 4 man-hours to cultivate an acre of corn. The profit from an acre of wheat is Rs. 2000 and from an acre of corn is Rs. 1500. How many acres of each crop should she plant to maximize profit?

Let x be the number of acres of wheat grown. Let y be the number of acres of corn grown.

The objective function, to be maximized, is given by:

Z = 2000x + 1500y

The constraints are:

6x + 4y <= 1200 (labor) 

x + y <= 100 (land) 

x >= 0, y >= 0

Solving the above system of linear inequalities graphically, we get the feasible region, as shown in the figure below.

The corner points of the feasible region are (0,0), (0,100), (200,0), and (100,50).

Computing the value of the objective function at each corner point, we get:

Z(0,0) = 0 Z(0,100) = 150000 Z(200,0) = 400000 Z(100,50) = 350000

Therefore, the maximum profit of Rs. 400000 is obtained by growing 200 acres of wheat and no corn.