3. (20 marks) A linear optimization model has been made for a metal fabricating shop where the variables are the number of screws, nuts, and bolts made per minute, de- noted as x1, x2, and x3 respectively. The objective function is in dollars; for example each screw gives a profit of $0.17 (or 17 cents) each. There are four operations: milling; cutting; lathework; and packaging. The algebraic formulation is: maximize 0.17x +0.09x2 +0.13x3 subject to milling 2x1 + 3x2 + 5x3 0 (a) Solve this model in LINGO or the Excel Solver, and obtain the sensitivity re- port. (b) Checking the allowable ranges, determine what would happen to the objective function value in each of the following situations (considered independently). (Assume that all proposed changes are feasible.) (i) The price of each nut rises by $0.15. (ii) An extra 100 units of milling becomes available. (iii) The number of units of lathework increases by 288, but the number of units of milling falls by 40. (iv) The price per bolt falls by $0.05. 3. (20 marks) A linear optimization model has been made for a metal fabricating shop where the variables are the number of screws, nuts, and bolts made per minute, de- noted as x1, x2, and x3 respectively. The objective function is in dollars; for example each screw gives a profit of $0.17 (or 17 cents) each. There are four operations: milling; cutting; lathework; and packaging. The algebraic formulation is: maximize 0.17x +0.09x2 +0.13x3 subject to milling 2x1 + 3x2 + 5x3 0 (a) Solve this model in LINGO or the Excel Solver, and obtain the sensitivity re- port. (b) Checking the allowable ranges, determine what would happen to the objective function value in each of the following situations (considered independently). (Assume that all proposed changes are feasible.) (i) The price of each nut rises by $0.15. (ii) An extra 100 units of milling becomes available. (iii) The number of units of lathework increases by 288, but the number of units of milling falls by 40. (iv) The price per bolt falls by $0.05