10.10 The Revenue-Maximizing Price
Once we know the demand equation, we can determine the revenue-maximizing price, after just a bit of algebraic manipulation and then one first derivative.
We start with the demand equation, as determined through the regression model.
Demand = 410.30 – 36.42x
Next, we will substitute demand for “Q” to represent “Quantity Demanded.”
Q = 410.32 – 36.42x
Then, we will substitute x for “P” to represent “Price,” since Price was our independent variable from the original model.
Q = 410.32 – 36.42P
Next, since Revenue = Price x Quantity, we can represent Q on the left side of our equation as R / P.
R / P = 410.32-36.42P
We can now multiply both sides of the equation by P in order to get rid of the fraction on the left side of the equation.
R = 410.32p – 36.42P^2
At this point, we have two variables and one equation. At first, it might seem as if we are stuck! However, we can take the first derivative of this equation, and then set its value to 0. The derivative represents a rate of change, and at the peak of the revenue function, the rate of change will (momentarily) be zero. Using the power rule, we will multiply 36.42 by the degree of the polynomial, and then reduce the polynomial’s degree by one. The result of this step is shown below:
R’ = 410.32 – 72.84P
Now, the only step left is to set R’ to 0, and then solve.
0 = 410.32 -72.84p
72.84P = 410.32
P = $5.63
We can also see this in a more manual way, by building a demand schedule based on our regression equation.
We can sort these values, as shown below.
Alternatively, we can view the revenue curve as a plot. Total revenue jumps up very slightly at first, as prices increase along the inelastic part of the curve. Later on, it falls as a result of the diminished demand.
Since Lobster Land has a fixed per-unit cost from its supplier ($3.50) we can add a profit column, as shown below, and then sort on that value.
The plot below depicts the expected profit at the various prices within our range.
