We have to maximize Z = 60x+ 15yFirst, we will convert the given inequations into equations, we obtain the following equations:x+y= 50, 3x+y= 90,x= 0 andy= 0

Region represented byx+y≤ 50:The linex+y = 50 meets the coordinate axes atA50,050,0andB0,500,50respectively. By joining these points we obtain the line 3x+ 5y= 15.Clearly0,00,0satisfies the inequationx+y≤ 50. So,the region containing the origin represents the solution set of the inequationx+y≤ 50.

Region represented by 3x+y≤ 90:The line 3x+y= 90 meets the coordinate axes atC30,030,0andD0,900,90respectively. By joining these points we obtain the line 3x+y= 90.Clearly0,00,0satisfies the inequation 3x+y≤ 90. So,the region containing the origin represents the solution set of the inequation 3x+y≤ 90.

Region represented byx≥ 0 andy≥ 0:Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequationsx≥ 0, andy≥ 0.

The feasible region determined by the system of constraints,x+y≤ 50, 3x+y≤ 90,x≥ 0, andy≥ 0, are as follows.



The corner points of the feasible region areO0,00,0,C30,030,0,E(20,30)E20,30andB0,500,50.

The values of Z at these corner points are as follows.

Corner pointZ = 60x+ 15yO0,00,060 × 0 + 15 × 0 = 0C30,030,060 × 30 + 15 × 0 = 1800E(20,30)E20,3060 × 20 + 15 × 30 =1650B0,500,5060 × 0 + 15 × 50 = 750

Therefore, the maximum value of Z is1800atthepoint(30,0)1800atthepoint30,0.Hence,x= 30 andy= 0 is the optimal solution of the given LPP.Thus, the optimal value of Z is 1800.