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When dealing with problems involving marginal revenue, it's crucial to understand the units of measure involved. In this exercise, production is measured in "hundred units," while revenue is measured in "thousand dollars."

This means that every time you sell 100 units of a product, you should measure the revenue in terms of thousands of dollars. It's like currency exchange for quantities and revenue values, and understanding this is essential before proceeding to any calculations involving marginal revenue.

The given marginal revenue, represented by the derivative \( R'(q) \), expresses the change in revenue per change in units produced, and uses these units to communicate that rate of change. So, in this particular problem, the unit of measure for marginal revenue is "thousand dollars per hundred units." This unit tells us how revenue increases or decreases when production adjusts by 100 units.

To interpret marginal revenue properly, you must read it in the context of these units. In this scenario where \( R'(16) = 0.15 \), this value indicates how much the revenue changes when the production and sale of goods increase incrementally.

What this actually means is quite simple: "If we increase the production and sales by 100 units, the revenue will increase by 0.15 thousand dollars."

To put this into perspective, 0.15 thousand dollars is equivalent to 150 dollars. Thus, selling additional 100 units yields an additional revenue of 150 dollars. It shows the economic impact of scaling up the production by providing a crucial insight into how sensitive revenue is to changes in the number of units sold.

The derivative \( R'(q) \) is central to the concept of marginal revenue, providing an application of calculus to economics. The derivative represents the rate of change of a function, and in this context, we use it to understand how revenue shifts as production quantity changes.

By calculating \( R'(q) \), businesses can predict how minor adjustments in production levels affect their revenue, enabling them to make informed decisions.

In this problem, \( R'(16) = 0.15 \) precisely quantifies the rate at which revenue increases when production and sales are marginally increased. This predictive power is invaluable for resource allocation, helping businesses determine the optimal level of production that maximizes profit without unnecessary expenditure.