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2.15
Colorful cupcakes
Seventeen cupcakes

Aunty Florrie's cupcakes

Aunt Florrie’s cupcakes were such a success at your cafe, that you decide to go into full time business selling her cupcakes state-wide! But you will have to make some important decisions to maximum your sales and profits.

In this step we will learn how the Demand and Supply curve analysis from Step 1.18 allows us to analyse profits by analysing a quadratic optimisation problem – an excellent application of our knowledge of the geometry of the parabola.

A tasty optimization problem

Suppose that every week your new expanded cupcake bakery has fixed costs of \(\normalsize{\$2000}\) and that it costs \(\normalsize{\$1}\) to make a dozen cupcakes.

How many dozen of Aunt Florrie’s cupcakes should you plan on making every week, and how should you price them? This is an optimization problem, with only one variable quantity, namely the number \(\normalsize{Q}\) of dozens of cupcakes you will make per week.

To get some initial data, you conduct two experimental week-long “bake sales” across your state:

  • in one week, you price the cupcakes at \(\normalsize{\$4}\) per dozen and sell \(\normalsize{3000}\) dozen cupcakes
  • in the other week, you price the cupcakes at \(\normalsize{\$8}\) per dozen and sell \(\normalsize{1000}\) dozen cupcakes.

Q1 (E): Which bake sale was most profitable?

Here is the Quantity / Price Demand Line that you deduce from this:

Coordinate plane with price on y axis and quantity on x axis; plot of line passing through points [1000,8] and [3000,4]

Notice that this is a different Demand Line than when we were selling individual cupcakes in your cafe: it is now the line through the points \(\normalsize{[1000,8]}\) and \(\normalsize{[3000,4]}\). It is also worth remarking that in Economics the axes are perhaps not what we might naively expect them to be: economists prefer to have Quantity along the horizontal axis, and Price along the vertical axis.

Q2 (E): How many dozen cupcakes would we expect to sell if the price was \(\normalsize{\$ 7}\)?

The Demand Line has slope

\[\Large{m=\frac{4-8}{3000-1000}=\frac{-4}{2000}=\frac{-1}{500}}\]

and so an equation of the form

\[\Large{P=-\frac{1}{500}Q+b}\]

for some number \(\normalsize{b}\). Please check that when we substitute one of the points, we determine that \(\normalsize{b=10}\), so that our Demand Line is

\[\Large{P=-\frac{1}{500}Q+10.}\]

Note that this equation could also be written as

\[\Large{Q+500P=5000 \quad \text {or as } \quad Q=5000-500P}. \tag 1\]

We see that the lower \(\normalsize{P}\) is, the higher \(\normalsize{Q}\) will be, and vice versa. Hopefully this makes sense: if we price our cupcakes lower, we can expect to sell more, and if we price them higher, we will sell less.

Coming to the crunch

Now how are we going to put this information together to actually make a business decision? If we sell \(\normalsize{Q}\) dozen cupcakes, priced at \(\normalsize{P}\) per dozen, then our Sales Income \(\normalsize S\) will be \(\normalsize{QP}\). In terms of the price \(\normalsize{P}\) which we have control over, this can be written, using (1), as

\[\Large{S=QP=(5000-500P)P}.\]

Q3 (E): What is our Sales Income if we sell cupcakes for \(\normalsize{\$ 4}\) per dozen? How about at \(\normalsize{\$ 6}\) per dozen? How about at \(\normalsize{\$ 8}\) per dozen?

Q4 (M): What is our Sales Income if we end up selling \(\normalsize{1500}\) dozen cupcakes?

We now want to maximize Sales Income, which we’ve seen is the quadratic function

\[\Large{S(P)=(5000-500P)P}\]

of \(\normalsize{P}\). Fortunately we know all about the shape of such functions–they are parabolas, and we know how to find the maximum values of a parabolic function!

Graph of S(P) as a function of P, with zeroes at [0,0] and [10,0] and maximum at x=5

This function will be zero when \(\normalsize{P=0}\) and when \(\normalsize{P=5000/500=10}\). So the maximum occurs when \(\normalsize{P}\) has the value exactly half way between the two zeroes, namely \(\normalsize{P=5}\). Plugging this into the expression for \(\normalsize{Q}\) and \(\normalsize{S}\), we find that

\[\Large{Q=2500}\]

and

\[\Large{S=QP=12,500}.\]

Our conclusions

To maximize Sales Income, we aim at selling \(\normalsize{Q=2,500}\) dozen cupcakes at \(\normalsize{P=\$5}\) per dozen, for an anticipated Sales Income of \(\normalsize{S=12,500}\).

Hopefully this shows you that maximization problems that occur in real life often revolve around the geometry of a quadratic relation, and so knowing about this geometry can help us solve practical problems.

Answers

A1. Profit from the first bake sale was \(\normalsize 4 \times 3000\) minus the cost of making \(\normalsize 3000\) dozen cupcakes minus the fixed cost:

\[\Large 4\times 3000 - 1 \times 3000 - 2000 = 7000.\]

Profit from the second bake sale was \(\normalsize 8 \times 1000\) minus the cost of making \(\normalsize 1000\) dozen cupcakes minus the fixed cost

\[\Large 8\times 1000 - 1 \times 1000 - 2000 = 5000.\]

So the first bake sale was more profitable.

A2. Visually from the diagram we see that \(\normalsize P=7\) occurs at about \(\normalsize Q=1500\).

A3. When \(\normalsize Q=4\) our formula gives \(S=\normalsize 3000\times 4 = 12000\) dollars. When \(\normalsize Q=6\) we get the same value of \(\normalsize S=12000\) dollars. If we check the value \(\normalsize Q=8\) we find that we’ve over-priced, and only gotten \(\normalsize S=1000\times 8 = 8000\) dollars in sales.

A4. If we want to express Sales in terms of Quantity we use

\[\Large{S=PQ=(-\frac{1}{500}Q+10)Q}.\]

Now if \(\normalsize Q=1500\) we get

\[\Large{S=10500}.\]

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This article is from the free online course:

Maths for Humans: Linear, Quadratic & Inverse Relations

UNSW Sydney