Boise Lumber has decided to enter the lucrative prefabricated housing business.

Question

Boise Lumber has decided to enter the lucrative prefabricated housing business. Initially, it plans to offer three models: standard, deluxe, and luxury. Each house is prefabdand partially assembled in the factory, and the final assembly is completed on site. The dollar amount of building material required, the amount of labor required in the factory for prefabrication and partial assembly, the amount of on-site labor required, and the profit per unit are as follows. Standard Model $6,000 Deluxe Model $8,000 Luxury Model $10,000 Material Factory Labor (hr)240220200 On-Site Labor (hr) Profit 300 S5,000 180 210 $3,400 $4,000 For the first year's production, a sum of $8,200,000 is budgeted for the building material; the number of labor-hours available for work in the factory (for prefabrication and partial assembly) is not to exceed 224,000 hr; and the amount of labor for on-site work is to be less than or equal to 243,000 labor-hours. Determine how many houses of each type Boise should produce to maximize its profit from this new venture. (Market research has confirmed that there should be no problems with sales.) standard model deluxe model luxury model houses houses houses

Explanation / Answer

This question is a simple question of linear equation.

let standard model houses to be built be X

let deluxe model houses to built be Y

let luxury model houses to be built be Z

the budget for materials is $8,200,000

maximum factory labor hour is $224,000

labor for on-site labor = $243,000

putting everything in 3 equations we can write as ......... (for material)

6,000X+ 8,000Y+ 10,000Z = 8,200,000 ......equation 1 => 3X+ 4Y+ 5Z = 4,100 (simplifying)

for factory labor

240X+ 220Y + 200Z = 224,000 ....equation 2 => 12X + 11Y+ 10Z = 11,200

for on-site labor

180X + 210Y + 300Z = 243,000 ....equation 3 => 6X + 7Y + 10Z = 8,100

multiplying equation 1 by 4 and then subtracting equation 2 with it we get

5Y +10Z = 5,200 ....equation 4

multiplying equation 3 by 2 and subtracting equation 2 we get

3Y+ 10Z = 5,000 ...equation 5

solving equation 4 & 5 we get

Z= 470

putting the value of Z in equation 4 or 5 we get the value of Y i.e

Y = 100

now putting the value of Y & Z in any of equations 1,2 or 3 we get the value of X i.e

X= 450

putting the value of X, Y&Z in profit equation we can get maximum profit

i.e 3400X + 4000Y + 5000Z = $4,280,000

so the standard models will be = 450 houses

deluxe models = 100 houses

luxury models = 470 houses.

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