A hand-drill manufacturer that produces two models, regular and portable, must assemble at least 100 drills for each day’s delivery. The regular drill requires 2 hrs of preparation and costs $10 and the portable requires 5 hrs of preparation and costs $15. The dealer’s objective is to minimize his total cost of preparation given that he has 400 preparation hrs available each day and at least 40 portables must be assembled each day.
2006-07-26 00:20:39 · 4 answers · asked by chizi 1 in Science & Mathematics ➔ Mathematics
x1 + x2 >= 100 ; x1 nr of regular / day, xx 2nr of portable / day
x2 >= 40 ; per day at least 40 poratblesssee
x1>=0 ; we dont steal
2x1 + 5x2 <= 400; at most 400 hours
min C(x1,x2) = 10*x1 + 15*x2
2006-07-26 00:45:49 · answer #1 · answered by gjmb1960 7 · 1⤊ 0⤋
If x-regulars, y- portables then
1. x+y>=100
2. 2x+5y =400
3. y>=40
4. x>=0
10x+15y must be minimized,
2006-07-26 07:41:21 · answer #2 · answered by Roxi 4 · 0⤊ 0⤋
Number of regulars is 'R' & no. of portables is 'P'
The equations are:
Minimise 10R+15P, Given that
P+R >= 100
2R+5P <=400
P>=40
R>=0
2006-07-26 07:38:28 · answer #3 · answered by mailtoneerajm 1 · 0⤊ 0⤋
dude, just make a system of equations
2006-07-26 07:32:24 · answer #4 · answered by x overmyhead 2 · 0⤊ 0⤋