Let required number of machine A and B are x and y respectively.

Since, products of each machine A and B are 60 and 40 units daily respectively. So, production by by x number of machine A and y number of machine B are 60x and 40y respectively.

Let Z denotes total output daily, so,

Z = 60x + 40y

Since, each machine of type A and B requires 1000sq. m and 1200 s. m area so, x machine of type A and y machine of type B require 1000x and 1200y sq. m area but,

Total available area for machine is 7600 sq. m. So,

1000x + 1200y 7600

or, 5x + 6y 38. {First Constraint}

Since each machine of type A and B requires 12 men and 8 men to work respectively. So, x machine of type A and y machine of type B require 12x and 8y men to work respectively.

But total men available for work are 72.

So,

12x + 8y 72

3x + 2y 18 {Second Constraint}

Hence mathematical formulation of the given LPP is,

Max Z = 50x + 40y

Subject to constraints,

5x + 6y 38

3x + 2y 18

x,y 0 [Since number of machines can not be less than zero]

Region 5x + 6y 38: line 5x + 6y = 38 meets the axes at A(,0), B(0,) respectively.

Region containing the origin represents 5x + 6y 38 as origin satisfies 5x + 6y 38

Region 3x + 2y 18: line 3x + 2y = 18 meets the axes at C(6,0), D(0,9) respectively.

Region containing the origin represents 3x + 2y 18 as origin satisfies 3x + 2y 18.

Region x,y 0: it represents the first quadrant.

Shaded region represents the feasible region.

The corner points are O(0,0), B(0,), E(4,3), C(6,0).

Thus the values of Z at these corner points are as follows:

The maximum value of Z is 360 which is attained at E(4,3), C(6,0).

Thus,the maximum output is Rs 360 obtained when 4 units of type A and 3 units of type B or 6 units of type A and 0 units of type B are manufactured.