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Solving Problems by Graphing Quadratic Functions

Module by: Kenny M. Felder. E-mail the author

Summary: This module covers how to solve problems through the graphic of 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.

Example 1

Problem 1

If a company manufactures xx size 12{x} {} items, its total cost to produce these items is x310x2+43xx310x2+43x size 12{x rSup { size 8{3} } - "10"x rSup { size 8{2} } +"43"x} {}. How many items should the company make in order to minimize its average cost per item?

Solution

“Average cost per item” is the total cost, divided by the number of items. For instance, if it costs $600 to manufacture 50 items, then the average cost per item was $12. It is important for companies to minimize average cost because this enables them to sell at a low price.

In this case, the total cost is x310x2+43xx310x2+43x size 12{x rSup { size 8{3} } - "10"x rSup { size 8{2} } +"43"x} {} and the number of items is xx size 12{x} {}. So the average cost per item is

A ( x ) = x 3 10 x 2 + 43 x x = x 2 10 x + 43 A ( x ) = x 3 10 x 2 + 43 x x = x 2 10 x + 43 size 12{A \) \( x \) = { {x rSup { size 8{3} } - "10"x rSup { size 8{2} } +"43"x} over {x} } =x rSup { size 8{2} } - "10"x+"43"} {}
(1)

What the question is asking, mathematically, is: what value of xx size 12{x} {} makes this function the lowest?

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

A ( x ) = x 2 10 x + 25 + 18 = x 5 2 + 18 A ( x ) = x 2 10 x + 25 + 18 = x 5 2 + 18 size 12{A \( x \) =x rSup { size 8{2} } - "10"x+"25"+"18"= left (x - 5 right ) rSup { size 8{2} } +"18"} {}
(2)

The graph opens up (since the x52x52 size 12{ left (x - 5 right ) rSup { size 8{2} } } {}term is positive), and has its vertex at 5,185,18 size 12{ left (5,"18" right )} {} (since it is moved 5 to the right and 18 up). So it would look something like this:

Figure 1
Graph of average price per item vs. number of items

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 xx size 12{x} {} items, the graph shows what A(x)A(x) size 12{A \( x \) } {}, the average cost per item, will be.

The point 5,185,18 size 12{ left (5,"18" right )} {} is the lowest point on the graph. It is possible for xx size 12{x} {} to be higher or lower than 5, but it is never possible for AA size 12{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.

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