Mathematics in Pizzeria

In a pizzeria, as everywhere else in the market place, the ability to count change and evaluate tips proves a handy skill. In a pizzeria, due to the nature of their main product, basic geometric knowledge is also an asset.

Say, in a near-by outfit, they currently offer one large pizza for the combined price of one small and one medium product of identical quality and design. Which should one choose? You of course remember the formula for the area of the circle, pr2, where p is the famous universal constant and r is the radius of the circle. If only you could measure their radii you would naturally choose the variant that provides more of the area, since the heights of the pizzas are identical and, therefore, irrelevant.

Unfortunately, you left the measuring tape at home, none is available in the pizzeria, and the only thing the tapping and searching of your pockets has produced is a protractor. Where did it come from? How did it get into the back pocket of your pants? You've no idea. But this is the only tool you have. It'll have to do. So, which one do you choose? Large or a combination of small + medium?

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    Mathematics in Pizzeria

    An ingenious approach to that quandary is described in 777 Mathematical Conversation Starters by John dePillis, p. 250.

    Ask the staff in the pizzeria to cut each of the three products in half. (The assumption here is that, since they know how to cut a pizza into 8 slices, they must be able to cut it into just two.) This yields three straight line segments -- the diameters of the pizzas. Make a triangle out of the three segments and use the protractor to measure its angles. If the triangle is acute, the combination small + medium gives you more cheese for your buck than one large pizza. The opposite is true if the triangle is obtuse. This is a direct consequence of the Law of Cosines. If, per chance, the triangle is right angled, then the Pythagorean theorem applies: area-wise it does not matter which one you choose, large or small + medium. (Think here of one of Euclid's proofs of the Pythagorean theorem.)

    In the latter case, an additional factor may tip the balance of your judgement. Do you like crust? In which case do you get more of it? With one large pizza or with a combination of small + medium of equal area?

    Assume the radii of the pizzas are (in common notations) rL, rM, and rS. From the equality of the areas,

    rL2 = rS2 + rM2 ≤ (rS + rM)2,

    so that, if the areas are equal, the circumferences are not: that of the larger pizza is shorter than the combined circumference of the small and medium:

    rL ≤ rS + rM,

    with the equality only when the small pizza is really too small. (You must recall the formula 2pr for the circumference.)

    In any event, a partial answer to the problem is that under certain circumstances a combination small + medium is preferred to one large pizza.

    Now, let's consider a different problem. What matters now is to really get more pizza for your buck, i.e., per (your) buck. Assume the areas of the pizzas are AL, AM, AS, while the price is PL, PM, and PS, respectively. "Pizza per buck" is measured by the ratio A/P of the area to the price. If AL/PL = AM/PM = AS/PS, we may proceed directly to the next important question, but if the "pizza per buck" ratio differs with the size, we may legitimately think of making a purchase that optimizes that ratio. Here is then a question: is it possible that a combination of two pizzas of different sizes optimizes the ratio A/P?

    Further exploration


    1. J. dePillis, 777 Mathematical Conversation Starters, MAA, 2002

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    Is it possible that a combination of two pizzas of different sizes optimizes the ratio A/P?

    The answer is "No." Indeed, let there be two pizzas with areas A1 and A2 priced respectively P1 and P2, such that A1/P1 differs from A2/P2. Is it possible that

    (A1 + A2)/(P1 + P2) > A1/P1 and also (A1 + A2)/(P1 + P2) > A2/P2?

    We know that it is not possible. If, for example, A1/P1 < A2/P2, then necessarily

    A1/P1 < (A1 + A2)/(P1 + P2) < A2/P2.

    In other words, to get the best deal, determine which pizza size has the largest ratio A/P and have as much of it as you can. No combinations here.

    How much pizza one can eat is a very personal question that is better left out of discussion. However, realistically, we can assume that monetary resources of a person in a regular pizza parlor are limited. This being the case, and assuming that one only buys whole pizzas, might this happen that a combination would now offer a better deal than a "pure" size selection?

    This is an example of a constrained optimization problem (one may realistically run in at a pizzeria.) Given that the resources are limited, but the person is willing to spend as much as it takes, is it possible that a combination of two pizzas of different sizes optimizes the ratio A/P?


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    Given that the resources are limited, is it possible that a combination of two pizzas of different sizes optimizes the ratio A/P?

    The answer is "Yes, quite." The situation is very different from the previous question.

    For simplicity assume there are just two kinds of pizza: one of area 25 in2 @ $4, the other 80 in2 @ $12. You have the amount of $20 you are willing to spend in its entirety. There are just two ways to spend $20. You may buy either five $4 pizzas or two $4 ones and one $12 pizza. In the first case, you'd get 125 in2 of pizza for $20, so that the A/P ratio equals 125/20 = 6.25. In the second case, 130 in2 of pizza for the same $20, and A/P in this case equals 130/20 = 6.5. The optimal ratio, that we would have used in the previous problem, is 80/12 (or 6 2/3), which is more than 6.6 and is thus better than any of the above, but does not apply, since the left over $8 would suffice to purchase more pizza.

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