Cost, Revenue and Profit Functions
Profit function
The gross profit is a simple function of the difference between the revenue obtained from the sale of a number of products and the costs involved in their production. Thus, Profit (P) = Revenue (R) – Costs (C).
For example Profit from x items = Revenue from x items – Cost of x items. Both Cost and Revenue are functions of x (number of items demanded or supplied). So, the Profit function is given by: P(x) = R(x) – C(x) where:
- x is the quantity of items demanded (supplied or produced)
- P(x) is the profit function in terms of x
- R(x) is the revenue function in terms of x
C(x) is the cost function in terms of x Note that the demand is assumed to be identical to supply unless otherwise stated. The profit point can be found by solving the equation: dP/dx = 0 for x where:
- x is the quantity of items demanded (supplied or produced)
- P is the profit function in terms of x
Example: A manufacturer knows that if x (hundred) products are demanded in a particular week, the total cost function is 14 + 3x and the total revenue function ($ 000) is 19x - 2x²
a) Derive the total profit function.
b) Find the profit break even points.
c) Calculate the level of demand that maximises profit (maximum profit point) and the amount of profit obtained.
Solution:
a) Total Profit function is given by:
P (x) = R(x) – C (x)
P(x) = 19x - 2x² – (14+3x)
P(x) = 16x - 2x² - 14
b) The profit break even points are the levels of demand which make
P(x) =0 when 16x - 2x² – 14 = 0. Solving this quadratic equation gives x =1 and x =7. Therefore profit break even points are when x = 100 and when x =700
c) Profit is maximised when dP/dx = 0. The dP/dx of 16x - 2x² – 14 gives 16 – 4x. And putting 16 – 4x =0 yields x = 4
Note: d²P/dx²= -4 (<0) showing that x = 4 gives the maximum profit.
The maximum profit when x = 4 is: P = 16 (4) – 2(4)² – 14. This yields 18. The weekly demand is 400 and the maximum profit is $18000.
Cost Function
The costs involved in standard processes can normally be categorised as follows:
a. Fixed (set-up) costs—They are associated with the purchase, rent or lease of equipment and fixed overheads. They are independent of number of items produced.
b. Variable costs—They are normally associated with supply of raw materials and overheads necessary to manufacture each product. They depend on the number of items produced.
c. Special costs—They are optional and might cover storage, maintenance or deterioration costs.
-They are usually small in value.
The total Cost function is given by: C (x) = a + bx +cx² where
- x is the quantity of items demanded (supplied or produced)
- a is the fixed cost associated with the product
- b is the variable cost per item
- c is the (optional) special cost factor
Example: The variable costs associated with a certain process are $ 0.65 per item. The fixed costs per day have been calculated as $ 250 with special costs estimated as $ 0.02x² where x is the number of items produced (size of the production run).
a) Derive a function to describe cost per item for a day’s production
b) Calculate the size of the daily run that will minimise cost per item
c) Find the cost of a day’s production for a run that minimise cost per item
Solution:
a) Total cost function: TC = 250 + 0.65x + 0.02x².
The cost per item, C = (250 + 0.65x + 0.02x² ) / x. This gives: 250/x + 0.65 + 0.02x
b) The cost per item is minimised when dC/dx = 0. The dC/dx of 250/x + 0.65 + 0.02x gives -250x^- 2 + 0.02. And putting -250x^-2 + 0.02 = 0 yields x = 112 (the size of the daily run that minimises cost per item).
c) The cost of a day’s run which minimises cost per item is given by: 250 + 0.65 (112) + 0.02 (112)² and yields $573.68
Revenue and Price Function
The general form of a Revenue function is given by: R(x) = x . pr(x) where:
- x is the quantity of items demanded (supplied or produced)
- pr(x) is the price function associated with the product
Price functions are used to vary the price of an item according to how many items are being considered. The more the number of items, the less the price per item and vice versa. This is based on the economy of scale. The general form of a price function is given by: pr(x) = a + bx where:
- x is the quantity of items demanded (supplied or produced)
- pr(x) is the price function
- a, b are coefficients that can take any numeric value.
Example: Determine the linear price function given that the price of an item is 3.50 when 250 items are demanded, but when only 50 items are demanded, the price rises to 5.50 per item. Calculate also the price per item at a demand level of 115.
Solution
pr(x) = a + bx and
- Substituting for pr = 3.5 and x = 250 gives: 3.5 = a + 250b
Substituting for pr = 5.5 and x = 50 gives: 5.5 = a + 50bSolving the two equations simultaneously yield b = -0.01 and a = 6
Therefore pr = 6 – 0.01x
- At demand level of 115, pr = 6 – 0.01 (115) which yields 4.85 per item
Example: Given the price function pr = 10.4 – 1.3x (where x is in hundreds), find the level of production which maximises total revenue.
Solution
R(x) = x . pr(x)
R = x . (10.4 - 1.3x)
•R = 10.4x – 1.3x²
Revenue is maximised when dR/dx = 0. The dR/dx of 10.4x – 1.3x² gives 10.4 – 2.6x
When 10.4 – 2.6x = 0, x = 4.
Thus the level of production necessary to maximise revenue is 400
Marginal Cost and Marginal Revenue Functions
If for some processes, the total cost function, C, and the revenue function, R, are identified, where x is the level of activity, the dC/dx is the marginal cost function (the extra cost incurred of producing another item at activity level x). The dR/dx is the marginal revenue function (extra revenue obtained from producing another item at activity level x). The maximum profit point can be found by solving the equation: dR/dx = dC/dx where C is the total cost function and R is the total revenue function.
Example: The fertiliser manufacturer can sell all bags of fertilisers of a particular type that they can produce. The total cost of producing q bags of fertiliser per week is given by 300q + 200. The price function is estimated as 500 – 2q.
a) Derive the revenue function, R.
b) Obtain the total profit function.
c) How many units per week should be produced in order to maximise profit?
d) Show that the solution of the equation dR/dq = dC/dq gives the same value for q as in part c.
e) What is the maximum profit available?
Solution
a) Revenue function, R = (500-2q)q. Thus R = 500q - 2q²
b) Total Profit function, P = R – C = 500q - 2q² –(300q – 2000). Thus P = 200q - 2q² – 2000
c) Profit is maximised when dP/dq = 0. The dP/dx of 200q - 2q2 – 2000 gives 200 – 4q. When 200 - 4q = 0 it yields q = 50. Therefore 50 units should be produced per week to maximise profit.
d) dR/dq = 500 – 4q and dC/dq = 300. So if dR/dq = dC/dq, it gives 500 – 4q = 300 which yields q = 50 units per week. This result agrees to answer in part C.
e) The maximum profit is found by substituting 50 units into the profit function, P = 200q - 2q² – 2000.
P = 200(50) - 2(50)² – 2000. Thus P = 3000
The revenue and cost functions can also be identified by using Integration technique. Integration is the reverse process to differentiation. For example, Integrating 6x2 gives 2x3 + C; Integrating 6x -10 gives 3x² - 10x + C. And given y when x =0 we can calculate the value for C. Now, the integration of the Marginal Revenue (MR) gives a Revenue function {R(x)}. The integration of the Marginal Cost (MC) gives a Cost function {C(x)}.
Example: The total revenue obtained (in MK 000) from selling x hundred items in a particular day is given by R, which is a function of variable x. Given that dR/dx = 20-4x:
a) Determine the total revenue function R(x).
b) Find the number of items sold in one day that will maximise the total revenue and evaluate this total revenue.
Solution
a) Integrating 20 – 4x gives 20x - 2x2 + C but when x = 0, R is also (=) 0. Therefore R(x) = 20x - 2x²
b) The value of x that maximises Revenue is found by solving equation dR/dx = 0. That is 20 – 4x = 0; x = 5. The Maximum revenue is R (5) = 20(5) – 2(52 ) = 50, 000.