Which is which?

[johnjones]

Sometimes, I'm confused and I find the wrong one...
I'm not sure when to find dp/dx or dx/dp. If the question told me specifically to find dp/dx, I could do it. But... when it's a wordy problem

i.e. Demand for a good is given by p = 100e^(-0.00125x). ($p/unit, x units/mo.)

(a) If price is currently $30.00/unit but is decreasing at the rate $1.50 per month, at what rate is demand increasing right now?
(b) Determine the unit price and monthly sales that will maximize revenue.

If I use dp/dx, I get 100e^(-0.00125x) * (-0.00125) {Chain Rule}
Then, I can use the fact that the first part of the expression is equal to p.

dp/dx = -0.00125p.

Now what if I use dx/dp? But quantity is also changing with respect to price? I'm so confused. :?:

 

[pka]

You have confused yourself and even confused me.
You could write x=[ln(p)-ln(100)]/(−0.00125).
Is that what you are trying to do?

 

[stapel]

I'm not sure when to find dp/dx or dx/dp.

Look at what they mean. If we assume that "p" is "price per unit" and "x" is "number of units", then "dp/dx" means something like "the change in the price as the production levels change" and "dx/dp" means something like "the change in production levels as the market price changes".

Note: You have defined "p" as being "demand", not "price".

(a) If price is currently $30.00/unit but is decreasing at the rate $1.50 per month...

You are given that the price is changing. However, you have defined no variable for "price".

...at what rate is demand increasing right now?

Until you provide a relationship between "price" and "demand" (and perhaps also "production levels"), I don't see how this can be answered.

(b) Determine the unit price and monthly sales that will maximize revenue.

Note that "revenue" is "income", found by multiplying the units sold ("x", I think) by the price per (for which, it seems, we have no variable).

Please reply with clarification, keeping in mind that math teachers might not be familiar with industry-specific (in this case, economics) terminology. Thank you.

Eliz.

 

[johnjones]

In economics, price is usually on the y-axis. Contrary, in math, for my course, price is on the x axis, and quantity (x) demanded is on the y-axis. Revenue is x times the demand equation. I have no clue about 'pka'... :?:

 

[stapel]

...for my course, price is on the x axis, and quantity (x) demanded is on the y-axis. Revenue is x times the demand equation.

Okay, but what is your variable for "price", and how does this relate to "demand"?

I have no clue about 'pka'.

He's just another tutor who is trying to figure out what you're saying.

Eliz.

 

[pka]

I have team-taught mathematical economics.
Revenue is always price times demand.

Price can be a function of demand.

Demand can be a function of price.


If p=100e<SUP>−0.00125x</SUP> where x is demand, the price is a function of demand.
But take the inverse function, x=[ln(p)-ln(100)]/(−0.00125), now we have demand as a function of price.

In the first case revenue is R(x)=[100e<SUP>−0.00125x</SUP>](x), a function of demand.

In the second case revenue is R(p)= {[ln(p)-ln(100)]/(−0.00125)}(p), a function of price.

 

[Gene]

I'll take a shot at it too. When it says

Demand for a good is given by p = 100e^(-0.00125x). ($p/unit, x units/mo.)

the equation is backwards. The independent variable is demand (x), the dependent is price (p). That's what PKA was attemptingto fix.
When they ask

(a) If price is currently $30.00/unit but is decreasing at the rate $1.50 per month, at what rate is demand increasing right now?

They are asking if dp ((\(\displaystyle \Delta\)$/unit) /mo) = -1.50 what is dx (\(\displaystyle \Delta\)units/mo).
You want dx/dp because that is \(\displaystyle \Delta\)(units/mo)/\(\displaystyle \Delta\)(($/unit)/mo)
You have dp/dx = -0.00125p so
dx/dp = 1/(dp/dx) =
-1/(0.00125p)
dx = -1/(0.00125*30)*dp =
-80/3*(-1.5) = 40 units/mo corrected
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Gene