Is there a correct way to include dimensions in economics equations?

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The discussion centers on the dimensional correctness of the demand function represented by the equation y=5000-p, where y denotes quantity demanded and p signifies price. Participants clarify that an implicit coefficient of 1.0 unit-of-price-1 is necessary for dimensional accuracy. They also illustrate how to express the equation with specific units, such as litres of water and items, ensuring that both sides of the equation maintain consistent dimensions. The final formulation presented includes units of demand and price, demonstrating the importance of dimensional analysis in economic equations.

PREREQUISITES
  • Understanding of demand functions in economics
  • Familiarity with dimensional analysis
  • Knowledge of unit conversion in economic contexts
  • Basic algebraic manipulation skills
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  • Research dimensional analysis in economics
  • Study the implications of unit consistency in mathematical modeling
  • Explore the use of implicit coefficients in economic equations
  • Learn about different units of measurement in economic contexts
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Economists, students of economics, and anyone involved in mathematical modeling of economic functions will benefit from this discussion, particularly those focusing on dimensional analysis and unit consistency in equations.

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A typical demand function is y=5000-p where y is quantity demanded and p is price. But this equation isn't dimensionally correct. What am I missing?
 
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Alexmahone said:
A typical demand function is y=5000-p where y is quantity demanded and p is price. But this equation isn't dimensionally correct. What am I missing?

Hi Alexmahone,

There's an implicit coefficient of $p$ that is apparently chosen to be $$1.0 \text{ unit-of-price}^{-1}$$, which will make it dimensionally correct.
 
I like Serena said:
Hi Alexmahone,

There's an implicit coefficient of $p$ that is apparently chosen to be $$1.0 \text{ unit-of-price}^{-1}$$, which will make it dimensionally correct.

Is this what you mean?

Assuming that price is to be measured in \$, and output in litres of water, the equation would be

[math]\frac{y}{1\text{ litre of water}}=5000-\frac{p}{$1}[/math]
 
Last edited:
Alexmahone said:
Is this what you mean?

Assuming that price is to be measured in \$, and output in litres of water, the equation would be

[math]\frac{y}{1\text{ litre of water}}=5000-\frac{p}{$1}[/math]

Basically, yes.
I actually left out the unit of demand, which is presumably a dimensionless number.

Then again, suppose the unit of demand is 1000 items and the unit of price is 10000 \$, then the formula would be:
$$y=5000 \text{ kItems} - 1 \frac{\text{kItems}}{10\text{ k}\$} \cdot p$$
 

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