Answer 1Let:x = pollutant reduction from factory X.y = pollutant reduction from factory Y.z = pollutant reduction from factory Z.Objective Function:Minimize Z = 15x + 10y + 20zZ - 15x - 10y - 20z = 0Eq. 1Subject to Constraints:TOC: 0.1x + 0.2y + 0.4z >= 30TON: 0.45x + 0.25y + 0.3z >= 40x,y,z >= 0Subtract e1 and e2 from L.H.S of constraint equations because inequality sign is ""0.1x + 0.2y + 0.4z - e1 = 30Eq. 20.45x + 0.25y + 0.3z - e2 = 40Eq. 3x, y, z , e1, e2 >= 01st IterationZxyze1e2RHSRatio Testrow 01-15-10-200000row 100.10.20.4-1030150row 200.450.250.30-1401602nd IterationZxyze1e2RHSRatio Testrow 0' = row 0 +10*row2'130-80-401600533row 1' = row 1-0.2*row2'0-0.2600.16-10.8-28row 2' = row2/0.2501.811.20-4160893rd IterationZxyze1e2RHSRatio Testrow 0'' = row0'/3010-30-13533-200row 1'' = row1'+0.26*row0''000-1-1-3137-256All Negative means can't be solved furtherrow 2'' = row2'-1.8*row0''-1016020-800-133Optimal Solution:x* = 533

Answer 2Supply in million gallons/day from the local reservoir = XSupply in million gallons/day from the pipeline = YObjective Function:Minimize Z = 300X + 500YSubject to Constraints:X+ Y >=10Eq.1X= 6Eq.3Y= 0Eq. 1Eq. 3XYXY01006100106Corner Points: A(4,6), B(5,6), C(5,10) and D(0,10)Eq. 2Eq. 4XYZXYXY46420051001056450050101051065000105000Optimal Solution: Z* = 4200; X* = 4; Y* = 6

Answer 2

Eq. 1Eq. 2Eq. 3Eq. 4

Ans 3A = $ invested in Project AB = $ invested in Project BObjective Function:Maximize Z = 0.1A + 0.15BSubject to:A+B = 24Eq.3X, Y >= 0Equation 1XY01260Equation 2XYSolving Eq.1 and Eq.3:04We get: Y = 6 and X = 3180Point B = (3,6)Solving Eq.2 and Eq.3:Equation 3Y = 2 and X = 9XYPoint C = (9,2)08120Corner points: A (0,12); B (3,6); C (9,2); D (18,0)XYZ012240Optimal Solution: Z* = 195; X* = 3 and Y* = 63619592265180450

= Feasible Region

Answer 6

Equation 1Equation 2Equation 3

= Feasible RegionOptimal Solution By Corner Point Method