In his book, Prophet of Innovation: Joseph Schumpeter and Creative Destruction, business historian Thomas K. McCraw lays out a numerical example of economies of scale. Unfortunately, in doing so he confuses marginal and average cost. McCraw writes (pp. 52-53):

Let us assume that a steel mill turns out ten pounds per day. Because the mill’s furnace must use a great deal of fuel to generate the heat needed to produce any steel at all, the cost of producing the first pound might be, say, \$10. But once the furnace is sufficiently hot, the mill’s operator can make additional steel by using just a little more fuel. So the average cost of producing the second pound might fall to \$9, the third to \$8, and so on, until the cost of the tenth pound is \$1. The average cost of all ten pounds would then be the sum of \$10 + \$9 + \$8 + \$7 + \$6 + \$5 + \$4 + \$3 + \$2 + \$1 (which equals \$55), divided by 10–that is \$5.50.

You can get an average cost of 10 pounds being \$5.50 if the marginal costs are \$10, \$9, etc., but then he should have said “marginal,” not “average.” Alternatively, he could have the cost of the first pound being \$10 and then additional pounds after that having a constant marginal cost of, say, \$5. Then the total cost of the 10 pounds would be \$10 + 9*\$5, for a total of \$55. In other words you have constant marginal costs from 2 pounds on and the declining average cost throughout that McCraw needs for his argument. Indeed, this second alternative is more fitting with his verbal reasoning. It makes sense for the first pound to have a big cost and then have additional units cost a constant amount each.