Crashing costs for a project are governed by the function:

50t2 + 100t
where t represents the number of days crashed (see graph). The project manager wants to crash the project, and is willing to pay up to $1,000 per day saved. How many days can be saved before the per-day cost is too high?

We can crash the project to the point where the per day cost of crashing equals $1,000, so the manager needs to find the value of t where the slope of the crash cost function equals that amount. The slope is given by the first derivative of the crash cost function:

Cost = 50t2 + 100t
= 2 × 50t2-1 + 100t1-1 = 100t + 100
1000 = 100t + 100
t = 9 days