Question

The owner of a business that makes and sells wood products intends to expand the work schedule by one half day each week and wants to optimize the use of that additional time. The firm makes five different items: a chair, a table, a desk, a bookcase, and a food-serving cart. The respective profits per unit are: $16, $30, $40, $42, and $32. The products require essentially the same basic operations: cutting, sanding and finishing, and assembly. The times for these operations differ for various items. However, the times are fairly standard. They are shown in the following table:

Time (minutes) per Operation

            Item                Cutting           Sanding & Finishing             Assembly

            Chair                    8                             12                                      4

            Table                    6                             10                                      3

            Desk                     9                             15                                      5

            Bookcase              9                             12                                      4

            Food Cart           12                             8                                      6

There are 320 minutes available for cutting, 400 minutes for sanding and finishing, and 270 minutes for assembling. What combination of products should be produced an the additional period each week in order to maximize profits? Clearly define your decision variables and formulate this problem as an LP problem.

Answer 1

let x1,x2,x3,x4,x5 denote the five different items such as chair,table,desk,bookcase and food cart.

the objective here is to maximize the objective of selling the product

z =16x1+30x2+40x3+42x4+32x5

constraints are the fulfillment of the weekly requirements of the various resources.

for cutting: => 8x1+6x2+9x3+9x4+12x5 <= 320

for sanding and finishing: => 12x1+10x2+15x3+12x4+8x5 <=400

for Assembly : => 4x1+3x2+5x3+4x4+6x5 < =270

decsion variables

x1,x2,x3,x4,x5 >= 0

library(lpSolve)

objective.in <-c(16,30,40,42,32)
const.mat <- matrix(c(8,6,9,9,12,12,10,15,12,8,4,3,5,4,6),nrow = 3,byrow = TRUE)
const.rhs = c(320,400,270)
const.dir = c("<=","<=","<=")
optimum <-lp(direction="max",objective.in,const.mat,const.dir,const.rhs)
optimum

Success: the objective function is 1413.333
optimum$solution

0.000000 0.000000 0.000000 31.111111 3.333333
optimum$objval

1413.333

which implies that bookcase = 31.11111 and food cart = 3.333 gives the maximum profit as 1413.333