In the modern businessworld, companies must find the most cost-efficient method of handling their inventory. One method is just-in-time inventory, where new inventory arrives just as existing stock is running out. As a simplified example of this, suppose that a heating oil company's terminal receives shipments of 8000 gallons of oil at a time and orders are shipped out to customers at a constant rate of 1000 gallons per day, where each shipment of oil arrives just as the last gallon on hand is shipped out. Inventory costs are determined based on the average number of gallons held at the terminal. So, how would we calculate this average?
To translate this into a calculus problem, let f(t) represent the number of gallons of oil at the terminal at time t (days), where a shipment arrives at time t = 0. In this case, f(0) = 8000. Further, for 0 < t < 8, there is no oil coming in, but oil is leaving at the rate of 1000 gallons per day. Since "rate" means derivative, we have f'(t) = -1000, for 0 < t < 8. This tells us that the graph of y = f(t) has slope -1000 until time t = 8, at which point another shipment arrives to refill the terminal, so that f(8) = 8000. Continuing in this way, we generate the graph of f(t) shown here at the left.
Since the inventory ranges from 0 gallons to 8000 gallons, you might guess that the average inventory of oil is 4000 gallons. However, look at the graph at the right, showing a different inventory function g(t), where the oil is not shipped out at a constant rate. Although the inventory again ranges from 0 to 8000, the drop in inventory is so rapid immediately following each delivery that the average number of gallons on hand is well below 4000.
As we will see in this chapter, our usual way of averaging a set of numbers is analogous to an area problem. Specifically, the average value of a function is the height of the rectangle that has the same area as the area between the graph of the function and the x-axis. For our original f(t), notice that 4000 appears to work well, while for g(t), an average of 2000 appears to be better, as you can see in the graphs.
Actually, several problems were just introduced: finding a function from its derivative, finding the average value of a function and finding the area under a curve. In this chapter, you will explore the relationships among these problems and learn a variety of techniques for solving them.