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Derivative : solution exercise 4.6 (Advanced math level)


4.6 Derivatives applied to biology, medicine and economy : Cost and revenue in economics, maximize your profit (marginal cost, marginal revenue)

In mathematical economics there is a relationship between revenue, r(x), and profit, p(x). The revenue is the money you get when you have produced (and thus sold) x items. The profit is calculated by substracting the costs, c(x), from revenues.

r(x) = revenues resulting from the production (and thus sale) of x items.

c(x) = costs resulting from the production of x items (workforce salaries, heating cost, costs of leasing your office space,...)

p(x) = r(x) – c(x) = profit due to the production (and thus sale) of x items.

You can observe that the revenue, the cost and the profit are all functions of the number of items produced, x. Also the derivative of the revenue with respect to x, r'(x), gives the marginal revenue. However, we know that a derivative is the calculation of a slope. Hence the marginal revenue r'(x) represents a In mathematical economics there is a relationship between revenue and profit. The revenue is the money you get when you have produced (and thus sold) x items. The profit is calculated by substracting the costs from revenues. The derivative of the revenue with respect to the items produced x is the marginal revenue (for an infinitesimal interval of r and of x), which gives the slope of the curve at various levels of production, which means for different values of x. So we know, thanks to the marginal revenue, what will be the increase in revenue, You can observe that the revenue, the cost and the profit are all functions of the number of items produced, x. Also the derivative of the revenue with respect to x, r'(x), gives the marginal revenue. However, we know that a derivative is the calculation of a slope. Hence the marginal revenue r'(x) represents the increase in revenue from producing (and thus selling) one more unit of a product, from producing (and thus selling) one more unit of a product (or one more item), You can observe that the revenue, the cost and the profit are all functions of the number of items produced, x. Also the derivative of the revenue with respect to x, r'(x), gives the marginal revenue. However, we know that a derivative is the calculation of a slope. Hence the marginal revenue r'(x) represents the increase in revenue from producing (and thus selling) one more unit of a product. The same goes for the derivative of the cost with respect to x, c'(x), which gives the marginal cost. The marginal cost is the increase in total cost as a result of producing one extra unit of product (or one extra item). The marginal cost is given for different production levels.

r'(x) = marginal revenue.

c'(x) = marginal cost.

You will find a graph hereunder that illustrates the important following fact:

The positive (= gain) maximum profit occurs at a production level (thus at a specific x value) where the marginal revenue is equal to the marginal cost :

p(x) = max if r'(x) = c'(x) and p(x) > 0

The positive maximum profit (gain) occurs at a production level (thus at a specific x value) where the marginal revenue is equal to the marginal cost

Note: in economics, the break-even point is the point at which there is no net loss or gain, thus cost and revenue are equal : c(x) = r(x).

Look at the graph hereabove: if a company produces more than 3,6 items per year, it is working above the break-even point and thus the profit is positive. It is a net gain for the company. A positive profit occurs when the revenue curve is above the cost curve thus
p(x) = r(x) – c(x) > 0. The company is earning money (net gain).
In our example (check on the graph), the positive maximum profit occurs at a production level greater than the break-even point, when both green tangent lines are parallel. This means that the maximum profit occurs when p'(x) = c'(x) – r'(x) = 0 (we find the x coordinate of the extrema using the first derivative of profit and looking for the value of x that gives a result that is equal to zero) thus when c'(x) = r'(x).

However, notice that at the far right of the graph the costs are greater than the revenues, this means that the company is loosing money. This might happen when a team is working inefficiently : in fact if the rate of production goes too high, operators must work overtime, machine maintenance cannot keep up, breakdowns occur, and the costs of production rise. Moreover, if this combines with a saturated market on which you cannot sell your product, the profit will sink.

To the left of the break-even point, the company is also loosing money. Costs are greater then revenues, the production is insufficient and revenues due to sales do not offset the high costs of production. The company is in a bad situation where
p'(x) = 0; c'(x) = r'(x) but p(x) is minimun and < 0. This is a situation where the profit is minimum and negative, thus the company is experiencing a maximum loss !

The maximum loss occurs for a level of production x where the green tangent lines are parallel too. However, in this case the situation is different from the maximum profit situation because here the cost curve is above the revenue curve : c(x) > r(x) thus
p(x) < 0 and this is a loss for the company.

You are asked to :

Given the following mathematical equations

r(x) = 10 . x

c(x) = x3 - 6.x2 + 15.x + 5

x = number of items produced, in thousands of units.

a) Is there a level of production for which the profit is maximum ? If your answer is yes, how many items must be sold to reach that level ?

p(x) = r(x) - c(x)
= (10x) - (x3 - 6x2 + 15x + 5)
= -x3 + 6x2 - 5x - 5

p'(x) = -3x2 + 12x - 5

If p'(x1) = 0 and p(x1) > 0 then x1 is the number of items that must be produced and sold to get the highest possible profit (or maximum profit).

You must now resolve the following quadratic equation : p'(x) = -3x2 + 12x - 5 = 0 to figure out the x-values that make that function equal to zero.

You get x = 0,47 or x = 3,53.

Now you must check if those two x values make the profit function p(x) positive.

p(0,47) = - 6,13 --> Negative value --> Loss!

p(3,53) = 8,13 --> Positive value --> Gain!

Conclusion, if you produce and sell 3,53 items (x = 3,53) then the first derivative of the profit p'(x) = 0 and p(x) > 0. This way you maximize your gains by making the revenues
r(x) greater than the costs c(x).

You must of course reject the solution x = 0,47 because it makes the profit negative generating a net loss of money due to r(x) < c(x) (revenues smaller than costs).

Note that x = 3,53 represents the production and sale of 3 items more 0,53 items. However it is impossible to produce and sell 0,53 items. Therefore you must choose to produce 3 or 4 items. Because you are the best CEO we know we will let that choice at your own discretion.

b) Find a level of production to minimize the mean costs of production.

The mean costs of production, are the costs generated to produce x items divided by the number of items produced.

Mean cost of production = c(x) / x = (x3 - 6.x2 + 15.x + 5) / (x)
= x2 - 6x + 15 + (5/x).

If you plot the graph of the mean cost of production with our free online graph plotter, you will observe that the function has a minimum at x = 3,24. This means that you found the optimal mean costs of production, which is the lowest mean cost, when you produce 3,24 items.

Pay attention to the fact that if you want to see something on the graph you traced, you need to trace it on the following range : x-axis : [0 ; 5] and y-axis : [0 ; 15]

c) The level of production that maximize the profit is different or equal to the level that minimize the mean costs of production ?

The level of production that maximize the profit is x = 3,53 items produced and sold while the level of production that minimize the mean cost is x = 3,24 items produced and sold. In fact those two values are very similar and in this exercise the point of maximum profit is very close from the point of minimum mean cost on the graph. However we observe here two distinct points that are clearly separated on the graph.

d) After you have answered the questions a), b) and c), use the free graph plotter online to trace r(x), c(x) and their respective derivative to check your answers and visualize graphically what is hiding behind those mathematical functions and mysterious derivatives.

I cannot do anything to help you on that question ! Trace it yourself with our online function tracer.


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