4.7 Solving problems by graphing quadratic functions

Surprisingly, there is a fairly substantial class of real world problems that can be solved by graphing quadratic functions.

These problems are commonly known as“optimization problems”because they involve the question:“When does this important function reach its maximum ?”(Or sometimes, its minimum ?) In real life, of course, there are many things we want to maximize—a company wants to maximize its revenue, a baseball player his batting average, a car designer the leg room in front of the driver. And there are many things we want to minimize—a company wants to minimize its costs, a baseball player his errors, a car designer the amount of gas used. Mathematically, this is done by writing a function for that quantity and finding where that function reaches its highest or lowest point.

In this case, the total cost is $x^3-\text{10}{x}^2+\text{43}x$ and the number of items is $x$ . So the average cost per item is

What the question is asking, mathematically, is: what value of $x$ makes this function the lowest ?

Well, suppose we were to graph this function. We would complete the square by rewriting it as:

The graph opens up (since the ${\left(x-5\right)}^2$ term is positive), and has its vertex at $\left(\mathrm{5,}\text{18}\right)$ (since it is moved 5 to the right and 18 up). So it would look something like this:

I have graphed only the first quadrant, because negative values are not relevant for this problem (why?).

The real question here is, what can we learn from that graph? Every point on that graph represents one possibility for our company: if they manufacture $x$ items, the graph shows what $A(x)$ , the average cost per item, will be.

The point $\left(\mathrm{5,}\text{18}\right)$ is the lowest point on the graph . It is possible for $x$ to be higher or lower than 5, but it is never possible for $A$ to be lower than 18. So if its goal is to minimize average cost, their best strategy is to manufacture 5 items, which will bring their average cost to $18/item.