A chain saw requires 7 hours of assembly and a wood chipper 3 hours. A maximum of 63 hours of assembly time is available. The profit of $190 on a chain saw and $220 on a chipper. How many of each should be assembled for maximum profit?

1 answer

number of chain saws --- x
number of chippers ----- y

7x + 3y ≤ 63 ---> time inequation

profit = 190x + 220y

we need integer values of x and y
only possible cases from 7x + 3y = 63 :
(0,21) , (3,14), (6,7), (9,0)

for (0,21) profit = 0 + 4620 = 4620
for (3,14) profit = 570 + 3080 = 3650
for (6,7) profit = 1140 + 1540 = 2680
for (9,0) profit = 1710+0 =1710

looks like they should make chippers only, they could make 21 of them for a profit of $4620.
Similar Questions
  1. solve this linear programming problem;A chain saw requires 4 hours of assembly and a wood chipper 6 hours. A maximum of 48 hours
    1. answers icon 2 answers
  2. solve this linear programming problem;A chain saw requires 4 hours of assembly and a wood chipper 6 hours. A maximum of 48 hours
    1. answers icon 3 answers
    1. answers icon 1 answer
    1. answers icon 1 answer
more similar questions