Thursday, August 17, 2006

Scaling of price of margarine

Often people think that physicists have to remember a lot of formulas like one for how to compute the resistance if you know the current and the voltage and another one for how to compute the voltage from the resistance and the current. If they are slightly more educated they realise that you only have to memorise R=U/I and algebra does the rest.

But actually, even that is not true. The way to think about Ohm's law is really to realise that for an Ohmian resistor the current is proportional to the voltage. And if you want, you can call the constant of proportionality resistance (or conductivity if you think in the opposite way). This is the important part of Ohm's law just like it's the 1/r^2 dependence of Newton's law (at least in 3D and at bit later you realise that this is just an expression of the analogue of Gauss' law) or that in string units the radius of the M-Theory circle is proportional to g_s (keeping alpha' fixed) as the mass of a D0 is proportional to 1/g_s. To know how things scale is enough in most cases rather than the knowledge of a formula.

So, let's apply this to an everyday situation. I am slightly worried about my weight so I want to buy Lätta margarine in the supermarket. It comes in two package sizes 250g and 500g. Let's take the prices from here, so you pay 0.85 Euro for 250g and 1.35 for 500g. Obviously, I buy the bigger package as I pay less than twice the money for twice the margarine.

But wait, can we compute how this price comes about? Let's assume the price consists of a price for the package and the price of the actual margarine. Of course, the price for margarine is proportional to the amount M of margarine. The price of the package is likely to be proportional to the surface of the margarine, so it scales like M^(2/3). Thus the total price is something like

P = M value + M^(2/3) package

Plugging in the two prices for the two sizes we can solve for "value" and "package". We find that the price of the margarine is -18.7 cents per kg. That's right, it has a negative price, just like for example nuclear waste. This opens up great possibilities, for example we can work out that 1.76 metric tons of margarine together with its package is exactly for free. Or, if I accept to take ten tons, Unilever will pay me 821.33 Euros! I see another get rich quickly scheme coming up.

10 comments:

John said...

A really pretty story!

DSB said...

Hi, You should have also added a constant peice that is independent of the margarine size to cover the supermarket costs since that is essentially independent of margarine size (not entirely but there is certainly a constant component). In fact that will be the greatest cost of the product to you. Hence you need to find another packet of different size to fix the third constant.

Robert said...

DSB (David?),

of course you are right. But if the price of that fixed (M^0) component is positive (which is probably save to assume) it only makes the bulk price more negative. Thus I was conservative by neglecting it.

Bee said...

You forgot the biggest contribution: advertisement cost. Doesn't scale with anything you actually get. I'd say it's actually quite scale independent.

So, what you really pay is the smile of the L├Ątta guy.

Best, B.

Chris Oakley said...

Robert,

I am trying to find a flaw in your logic, but I cannot.

Obviously there is a big commercial opportunity out there, and I think that you should just go for it.

Robert said...

I am quite embarrassed: My last comment was of course complete nonsense!!!

By allowing for a price component "fixed" which scales like M^0 (i.e. is constant), the per volume price of course goes up, not down as I claimed. More spcifically,

d value / d fixed = (2^(2/3)-1)/(2-2^(2/3)) > 0.

Kasper Olsen said...

Great fun and great use of elementary physics!

Cheers, Kasper

PS: If you do worry about your weight, you might not be tempted to get the 1.76 tons of magarine...

Jonathan Andrew, USA said...

You are amazing!

karneval said...
This comment has been removed by a blog administrator.
Bloody Rage said...

I am so glad to see this. Meat definitely needs to be cooked to a certain point. This is not good for you when it is not cooked to this point.