You can see that the marginal costs are constant at $65,000 per extra 1,000 units and this is simply the extra variable costs of $65 per unit.
Moving from 1,000 units to 2,000 units generates an extra $120,000 revenue for additional costs of $65,000. So, worthwhile.
Moving from 2,000 units to 3,000 units generates an extra $80,000 revenue for additional costs of $65,000. So, worthwhile.
Moving from 3,000 units to 4,000 units generates an extra $40,000 revenue for additional costs of $65,000. So, not worthwhile. Nor is it worthwhile increasing quantity any further.
When MR > MC, sell more. When MR < MC sell less. Profits are maximised when MC = MR.
Therefore, as before when we looked directly at profits, we have identified that sales of 3,000 units will be where profits are maximised and these sales will be generated if the selling price is set at $120.
The algebraic approach
The problem with the tabular approach used above is that it only looked at the data in increments of 1,000 units. That allowed us to conclude that profit was maximised when 3,000 units were sold. But could profits be better if, say, sales were 2,900, 3,010 or 3,100 units? We have no information in the table that allows the advice to be refined.
To obtain greater accuracy instead of relying on tables, the relationship between quantity and price has to be described by an equation. The equation will be of the form:
P = a - bQ
Where P = price per unit and Q = quantity sold. a and b are constants that have to be found using the demand curve data. This formula is supplied in the PM exam.
We know from the table that at a price of (say) $140, 2,000 units are demanded. So:
140 = a – b x 2,000
Moving to the previous quantity level, when Q = 1,000, P = 160. So:
160 = a – b x 1,000
a and b are then found by simultaneous equations. Subtracting the two equations gives:
-20 = -b x 1,000
So b = 20/1,000 = 0.02
Put that information into either of the two original equations to find a. So the first equation becomes:
140 = a – 0.02 x 2,000 = a – 40
a = 180.
The demand curve is therefore
P = 180 – 0.02Q
Note that the formula sheet gives a simple way of finding a and b if the required information is supplied:
a = price when Q = 0. From the table this is 180
b = change in price/change in quantity. For each increment of the table price changes by 20 and the quantity changes by 1,000, so b = 20/1,000 = 0.02
The formula sheet also shows how to convert the demand curve equation to an equation for the marginal revenue.
Demand curve: P = a – bQ P = 180 – 0.02Q
MR: MR = a – 2bQ MR = 180 – 0.04Q
We know that the marginal cost per unit is $65, the variable cost per unit. Profit is maximised when
MC = MR
65 = 180 – 0.04Q
Solving for Q:
-115 = -0.04Q
115 = 0.04Q
Q = 115/0.04 = 2,875
You will see that this is slightly less than our previous answer of 3,000. What’s happened is that as output increases from 2,000 to 3,000, the early additional units (2,001, 2,002…etc) are worthwhile but later ones are not (2,997, 2998, 2,999, 3,000). However, all of this detail is lost and in total moving from 2,000 to 3,000 is worthwhile. Moving from 2,000 to 2,875 is even better.
Now used this quantity figure in the original demand curve equation to work out the selling price.
P = 180 – (0.02 x 2,875)
P = 122.5
So, using the algebraic approach, profits will be maximised when 2,750 units are sold and these sales will be generated if the selling price is set at $122.50.
The second article on pricing will deal with more practical aspects of pricing decisions.
Written by Ken Garrett, a freelance lecturer and writer