A bicycle manufacturing company makes a particular type of bike. Each child bike requires 4 hours to build and 4 hours to test. Each adult bike requires 6 hours to build and 4 hours to test. With the number of workers, the company is able to have up to 120 hours of building time and 100 hours of testing time for a week. If c represents child bikes and a represents adult bikes, can the company build 10 child bikes and 12 adult bikes in a week
Let's first calculate the total time needed to build and test 10 child bikes:
10 child bikes x 4 hours/build + 10 child bikes x 4 hours/test = 40 + 40 = 80 hours for child bikes
Similarly, let's calculate the total time needed to build and test 12 adult bikes:
12 adult bikes x 6 hours/build + 12 adult bikes x 4 hours/test = 72 + 48 = 120 hours for adult bikes
Therefore, the total time needed to build and test 10 child bikes and 12 adult bikes is:
80 + 120 = 200 hours
However, the company only has 120 hours of building time and 100 hours of testing time available. This means that they cannot build both 10 child bikes and 12 adult bikes in a week.
Let's use c to represent the number of child bikes and a to represent the number of adult bikes. We can set up a system of inequalities to represent the constraints:
4c + 6a ≤ 120 (total building time cannot exceed 120 hours)
4c + 4a ≤ 100 (total testing time cannot exceed 100 hours)
We also need to specify that c and a must be non-negative integers (i.e. they cannot be fractions or negative numbers).
Using these constraints, we can check if the company can build 10 child bikes and 12 adult bikes:
4(10) + 6(12) = 88 ≤ 120 (this constraint is satisfied)
4(10) + 4(12) = 88 ≤ 100 (this constraint is also satisfied)
Since all the constraints are satisfied, the company can build 10 child bikes and 12 adult bikes in a week.