part a and b A company manufactures Products A, B, and C. Each product is processed in three departments: I, II, and III. The total available labor-hours per week for Departments I, II, and III are 1080, 1200, and 960, respectively. The time requirements (in hours per unit) and the profit per unit for each product are as follows:| | Product A | Product B | Product C ||--------------|-----------|-----------|-----------|| Dept. I | 2 | 1 | 2 || Dept. II | 3 | 1 | 2 || Dept. III | 2 | 2 | 1 || Profit | $18 | $12 | $15 |If management decides that the number of units of Product B manufactured must equal or exceed the number of units of products A and C manufactured, how many units of each product should the company produce to maximize its profit?- Product A: 120 (Incorrect)- Product B: 200- Product C: 320 (Correct)What is the maximum profit (in dollars)?- $8640 (Correct)

Given that a company manufactures products A,B, and C.Total available labor hours per week for department I, II, and III are 1080,1200 and 960 respectively.The time required and the profit per unit for each product is given below.ABCDept. I212Dept. II312Dept. III221Profit$18$12$15Given that the number of unit of products B must equal or exceed number of units of prodct A and C.Let x unit of product A, y unit pf product B and z unit of prodcut C is manufactured per week.Given that total available labor hours per week for department I, II, and III are 1080,1200 and 960 respectively.Therefore:Given that the number of unit of products B must equal or exceed number of units of product A and C.Therefore .Objective function is .Therefore adding slack variables, the standard form of LPP is written as follows.Subject to:First simplex table is created as follows.B.VxyzS1S2S3S4XBMin. Ratio XBx1S12null" style="cursor: default;" data-mce-style="cursor: default;">nullnull" style="cursor: default;" data-mce-style="cursor: default;">null1null" style="cursor: default;" data-mce-style="cursor: default;">nullnull" style="cursor: default;" data-mce-style="cursor: default;">nullnull" style="cursor: default;" data-mce-style="cursor: default;">null1080540S23null" style="cursor: default;" data-mce-style="cursor: default;">nullnull" style="cursor: default;" data-mce-style="cursor: default;">nullnull" style="cursor: default;" data-mce-style="cursor: default;">null1null" style="cursor: default;" data-mce-style="cursor: default;">nullnull" style="cursor: default;" data-mce-style="cursor: default;">null1200400S3null" style="cursor: default;" data-mce-style="cursor: default;">nullnull" style="cursor: default;" data-mce-style="cursor: default;">nullnull" style="cursor: default;" data-mce-style="cursor: default;">nullnull" style="cursor: default;" data-mce-style="cursor: default;">nullnull" style="cursor: default;" data-mce-style="cursor: default;">null1null" style="cursor: default;" data-mce-style="cursor: default;">null960480S41−11null" style="cursor: default;" data-mce-style="cursor: default;">nullnull" style="cursor: default;" data-mce-style="cursor: default;">nullnull" style="cursor: default;" data-mce-style="cursor: default;">null101→Zj−Cj−18↑−12−150000Observe that −18 is the most negative value in the Zj−Cj row, therefore x is incoming variable.Observe that 1 is the minimum ratio in the last column, therefore S4 is outgoing variable.Now the second iterative table is obtained by applying following operations.B.VxyzS1S2S3S4XBMin. Ratio S10301" data-mce-style="cursor: default;">" data-mce-style="cursor: default;">−21080360S204−1" data-mce-style="cursor: default;">1" data-mce-style="cursor: default;">−31200300S304−1" data-mce-style="cursor: default;">" data-mce-style="cursor: default;">1−2960240→x1−11" data-mce-style="cursor: default;">" data-mce-style="cursor: default;">" data-mce-style="cursor: default;">10−−Zj−Cj0−30↑300018Observe that −30 is the most negative value in the Zj−Cj row, therefore y is incoming variable.Observe that 240 is the minimum ratio in the last column, therefore S3 is outgoing variable.Now the third iterative table is obtained by applying following operations.B.VxyzS1S2S3S4XBMin. Ratio S1000.751" data-mce-style="cursor: default;">−0.75−0.5360480S200001−1−1240−−y01−0.25" data-mce-style="cursor: default;">" data-mce-style="cursor: default;">0.25−0.5240−−x100.75" data-mce-style="cursor: default;">" data-mce-style="cursor: default;">0.250.5240320→Zj−Cj00−4.5↑007.53Observe that −4.5 is the most negative value in the Zj−Cj row, therefore z is incoming variable.Observe that 320 is the minimum ratio in the last column, therefore x is outgoing variable.Now the third iterative table is obtained by applying following operations.B.VxyzS1S2S3S4XBMin. Ratio S1−1001" data-mce-style="cursor: default;">−1−1120S200001−1−1240yMath input errorMath input errorMath input errorMath input errorMath input errorMath input errorMath input errorMath input errorMath input errorMath input error0.33510" data-mce-style="cursor: default;">" data-mce-style="cursor: default;">0.340.34320z1.3401" data-mce-style="cursor: default;">" data-mce-style="cursor: default;">0.340.67320Zj−Cj6000096Since all the entries in the row Zj−Cj≥0, therefore optimality is reached.Therefore the maximum profit occurs at y=320,z=320,x=0.The maximum profit is .Hence the maximum profit occurs at x=0,y=320,z=320 with maximum profit $8640.