Maximize = 2X + 3Y subject to 2X + 3Y 300 X + Y 300 X + 3Y 240 X, Y 0 A. Formulate the Simplex tableau B. Identify the pivot column, pivot row and pivot element from the simplex table C. Determine the maximum values of X and Y that maximize the firms profit D. Calculate the maximum profit of the firm

We have, Let x units of food A and y units of food B Where x 2 0,y2 0 A Diet for a sick person must contain at least 4000 units of vitamins, 50 units of minerals and 1400 units of calories. Two foods A and B are available at a cost of Rs 5 ans Rs 4 per unit respectively. one unit of food A contains 200 units of vitamins, 1 units of minerals and 40 units of calories, while one unit of the food B contains 100 unit of vitamins, 2 unit of minerals and 40 units of calories We are to minimize z=5x +4y Subject to constraints 200 x + 100y 2 4000 (for vitamins) =>2x+y2 40....(1) x + 2y250 ......(2) (forminerals) 40 x +40 y 2 1400 (for calories) =>x+y> 35 .......(3) Where x 2 0 , y 2 0 From equation (1) we get, 2x+y=40 If x = 0 then y =40 OY in L (0, 40) If y = 0 then x = 20 OX in A (20, 0) From equation (2) we get, x + 2y=50 If x = 0 then y =25 OY in M (0 , 25) If y = 0 then x=50 OX in B (50 , 0) From equation (3) we get, xty=35 If x = 0 then y =35 OY in N (0 , 35) If y = 0 then x = 35 OX in C(35 , 0) L(0, 40) N(0. 35) E(5.30), M(O. 25) D(20. 15) B(50. 0) O A(20. 0) X (35. 0) Shaded region is the feasible region, which is unbounded. Then, Comer point is B ( 50, 0) D (20 , 15) E(5 , 30) L(0 , 40) At point B( 50 , 0) = Z=5 (50) +4 (0)=250 At point D (20,15) =>Z= 5(20)+4 (15)=100+60=160 At point E (5 , 30) => Z=5 (5) +4 (30)=25 +120=145 At point L (0 , 40) => Z=5 (0) + 4 (40) =0+160=160 Then at least cost = Rs 145 at (5 , 30 ) Hence, this is the