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I am working on a model to set a price that maximizes profits. The equation for profits is:

Profits=price x (# sold) - (fixed cost) x (# sold)

I have models that predict (# sold) and (fixed cost) is a constant. So my first thought is to take the derivative of Profits with regards to price and set it equal to 0:

dProfits/dprice = 0

but since profits is linear and (# sold) and (fixed cost) are basically constants in this scenario, wouldn't that just be:

dProfits/dprice = (# sold)

My calculus is a little rusty, so any tips are very much appreciated. Or if there's something I'm missing about the profit equation, or if there's some algorithm I could use in this situation.

user3476463
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  • If # sold were constant (and > 0), the optimal price would be infinity. But #sold is not constant, and presumably is a function of price (what is called the demand curve). Generally, demand (# sold) is a downward sloping function of price. However, the opposite can occur, to an extent, for a so-called Veblen good. – Mark L. Stone May 23 '18 at 15:03
  • @MarkL.Stone Thank you for getting back to me so quickly. I have forecasts that will predict the (# sold) for a particular day. So on a particular day, if I know the forecasted (# sold) I was trying to figure out how to set the price to maximize the profits. So are you suggesting I need to get (# sold) as a function of price instead of just using the output from the forecast? – user3476463 May 23 '18 at 15:50
  • What is your (current) forecast based on? Doesn't price affect forecast? Do you think you can sell the same number of ice cream cones for $1 trillion each as for $1 each? That is Economics 101. – Mark L. Stone May 23 '18 at 15:53
  • @MarkL.Stone Ok so you're suggesting don't treat the forecasted value of (# sold) for that day as a constant. Instead substitute the forecast formula for (# sold) in the profit equation and take the derivative. Is that correct? – user3476463 May 23 '18 at 16:08
  • Something like that. But derivative is only appropriate if the function is differentiable, and even then, not applicable if the optimum lies on a boundary (for instance, at quantity zero). – Mark L. Stone May 23 '18 at 16:12
  • @MarkL.Stone hm, I'm using an arima model so the formula is something like (# sold)=0.5 x lag1(# sold)+0.65 x lag2(# sold) + residuals. Do you think it makes sense to take the derivative in this case, or is there another approach you would suggest? – user3476463 May 23 '18 at 16:26
  • @MarkL.Stone hm, I'm using an arima model so the formula is something like (# sold)=0.5 x lag1(# sold)+0.65 x lag2(# sold) + 0.7 x lag1(price) + residuals. Do you think it makes sense to take the derivative in this case, or is there another approach you would suggest? – user3476463 May 23 '18 at 16:26
  • Now you have a dynamic multi-period (optimization) problem, so the optimization is more complicated. Price in current period will affect quantity in future periods. Then how to aggregate and discount future profits, etc. This may be beyond your capabilities. To really do it right, you should also account for the uncertainty in forecasts. – Mark L. Stone May 23 '18 at 16:31

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