Dimensional Analysis & Sales Tax

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Explain this!

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If dimensional analysis is used for the following, would the calculation appear as follows?

One television set for $250.00 with sales tax of 6%

$250 * 0.06/ TV set * 1 TV set = $15.00 or $250/TV set * 0.06 * 1 TV set = $15.00
 
Explain this! said:
If dimensional analysis is used for the following, would the calculation appear as follows?

One television set for $250.00 with sales tax of 6%

$250 * 0.06/ TV set * 1 TV set = $15.00 or $250/TV set * 0.06 * 1 TV set = $15.00
Click to expand...
What's the question?
 
Explain this! said:
If dimensional analysis is used for the following, would the calculation appear as follows?

One television set for $250.00 with sales tax of 6%

$250 * 0.06/ TV set * 1 TV set = $15.00 or $250/TV set * 0.06 * 1 TV set = $15.00
Click to expand...
If the question is, "what amount of sales tax do I have to pay?", I would just say $250 * 0.06 = $15.

On the other hand, if you are asked, "how much tax do you pay per set?", I would say $250/set * 0.06 = $15/set.

The difference is entirely in what the goal is. In the latter case, we are thinking of the price per set, while in the former we are focused just on this transaction. But ultimately, it doesn't really matter.

On the other hand, "0.06/set" doesn't really make sense, as the tax is not related to the number of sets, but to the dollar amount.
 
I think I went through this with you in a thread about a year ago. Dimensional analysis is impossible to do with this kind of problem unless you define your units very carefully. Furthermore, dimensional analysis is convenient only with formulas that involve only multiplication, division, and exponentiation; otherwise you need to worry about common denominators.

[math]\text {Let } \dfrac{f}{\text {unit}} = \dfrac{\text {final cost in dollars}}{\text {unit}}.[/math]
[math]\text {Let } \dfrac{p}{\text {unit}} = \dfrac{\text {pre-tax price in dollars}}{\text {unit}}.[/math]
[math]\text {Let } \dfrac{t}{\text {unit}} = \dfrac{\text {tax in dollars}}{\text {unit}}.[/math]
Do you see that we have three things denominated in dollars? You cannot just use a dollar sign for three different variables.

Here are our formulas

[math]\dfrac{f}{\text {unit}} = \dfrac{p}{\text {unit}} + \dfrac{t}{\text {unit}}.[/math]
[math]\dfrac{t}{\text {unit}} = \dfrac{0.06p}{\text {unit}}.[/math]
Now we can use dimensional analysis.

[math]\dfrac{f}{\text {unit}} = \dfrac{p}{\text {unit}} + \dfrac{t}{\text {unit}} = \dfrac{p}{\text {unit}} + \dfrac{0.06p}{\text {unit}} = \dfrac{1.06p}{\text {unit}} \implies[/math]
[math]\text unit * \left (\dfrac{f}{\text {unit}} = \dfrac{1.06p}{\text {unit}} \right ) \implies f = 1.06p.[/math]
Notice that f and p here are UNITS, not variables.

Now we it is simple to go [math]f = 1.06p \text { and } p = \$250 \implies f = 1.06 * \$250 = \$265.[/math]
So you can justify the result using dimensional analysis, but it is not a clean method as it is in physics or chemistry due to the confusion of different kinds of dollars and the use of addition and subtraction in financial formulas.
 
JeffM said:
I think I went through this with you in a thread about a year ago. Dimensional analysis is impossible to do with this kind of problem unless you define your units very carefully. Furthermore, dimensional analysis is convenient only with formulas that involve only multiplication, division, and exponentiation; otherwise you need to worry about common denominators.

[math]\text {Let } \dfrac{f}{\text {unit}} = \dfrac{\text {final cost in dollars}}{\text {unit}}.[/math]
[math]\text {Let } \dfrac{p}{\text {unit}} = \dfrac{\text {pre-tax price in dollars}}{\text {unit}}.[/math]
[math]\text {Let } \dfrac{t}{\text {unit}} = \dfrac{\text {tax in dollars}}{\text {unit}}.[/math]
Do you see that we have three things denominated in dollars? You cannot just use a dollar sign for three different variables.

Here are our formulas

[math]\dfrac{f}{\text {unit}} = \dfrac{p}{\text {unit}} + \dfrac{t}{\text {unit}}.[/math]
[math]\dfrac{t}{\text {unit}} = \dfrac{0.06p}{\text {unit}}.[/math]
Now we can use dimensional analysis.

[math]\dfrac{f}{\text {unit}} = \dfrac{p}{\text {unit}} + \dfrac{t}{\text {unit}} = \dfrac{p}{\text {unit}} + \dfrac{0.06p}{\text {unit}} = \dfrac{1.06p}{\text {unit}} \implies[/math]
[math]\text unit * \left (\dfrac{f}{\text {unit}} = \dfrac{1.06p}{\text {unit}} \right ) \implies f = 1.06p.[/math]
Notice that f and p here are UNITS, not variables.

Now we it is simple to go [math]f = 1.06p \text { and } p = \$250 \implies f = 1.06 * \$250 = \$265.[/math]
So you can justify the result using dimensional analysis, but it is not a clean method as it is in physics or chemistry due to the confusion of different kinds of dollars and the use of addition and subtraction in financial formulas.
Click to expand...

Thank you for the reply! It's somewhat complicated though.
 
Explain this! said:
Thank you for the reply! It's somewhat complicated though.
Click to expand...
That is exactly my point. Dimensional analysis works well with formulas that have clearly differentiated units and no additions or subtractions. In finance and economics, where we must distinguish between monetary units attached to different concepts, and where the formulas are not just concatenations of multiplications and divisions, dimensional analysis makes things super complex.

Long ago, I did read an essay by someone trained in the physical sciences trying to explain accounting to engineers. I suspect that some who had graduated from MIT, CalTech, Carnegie-Mellon, or Rice found it enlightening. Frankly, I think finance, accounting, and economics are best understood without trying to force them into analytic modes designed for the physical sciences.
 
You have to pay the price PLUS the tax. But the tax is 0.06 times the price.

[math]p + t = p + 0.06 * p = p * (1 + 0.06) = p * 1.06.[/math]
And that basic algebra is a lot easier to understand than messing around with dimensional analysis.
 
JeffM said:
I think I went through this with you in a thread about a year ago. Dimensional analysis is impossible to do with this kind of problem unless you define your units very carefully. Furthermore, dimensional analysis is convenient only with formulas that involve only multiplication, division, and exponentiation; otherwise you need to worry about common denominators.

[math]\text {Let } \dfrac{f}{\text {unit}} = \dfrac{\text {final cost in dollars}}{\text {unit}}.[/math]
[math]\text {Let } \dfrac{p}{\text {unit}} = \dfrac{\text {pre-tax price in dollars}}{\text {unit}}.[/math]
[math]\text {Let } \dfrac{t}{\text {unit}} = \dfrac{\text {tax in dollars}}{\text {unit}}.[/math]
Do you see that we have three things denominated in dollars? You cannot just use a dollar sign for three different variables.

Here are our formulas

[math]\dfrac{f}{\text {unit}} = \dfrac{p}{\text {unit}} + \dfrac{t}{\text {unit}}.[/math]
[math]\dfrac{t}{\text {unit}} = \dfrac{0.06p}{\text {unit}}.[/math]
Now we can use dimensional analysis.

[math]\dfrac{f}{\text {unit}} = \dfrac{p}{\text {unit}} + \dfrac{t}{\text {unit}} = \dfrac{p}{\text {unit}} + \dfrac{0.06p}{\text {unit}} = \dfrac{1.06p}{\text {unit}} \implies[/math]
[math]\text unit * \left (\dfrac{f}{\text {unit}} = \dfrac{1.06p}{\text {unit}} \right ) \implies f = 1.06p.[/math]
Notice that f and p here are UNITS, not variables.

Now we it is simple to go [math]f = 1.06p \text { and } p = \$250 \implies f = 1.06 * \$250 = \$265.[/math]
So you can justify the result using dimensional analysis, but it is not a clean method as it is in physics or chemistry due to the confusion of different kinds of dollars and the use of addition and subtraction in financial formulas.
Click to expand...
In you calculation, does "unit" represent any particular amount or number? I am beginning to understand your example, but just using "unit" as a denominatior confuses me.
 
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