- Problem
- Detailed Solution
- Summary Solution

### Marginal average profit

Let cost *C* and price *p*, both as a function of
quantity *x*, be given by

\(K(x)=0.06x^3-x^2+50x+400\), and

\(p(x)=150-0.4x\) .

Determine marginal average profit for output *x *= 40.

### Structure of the problem

In this exercise, the first derivative of average profit must be determined as a function of quantity x. That function then has to be evaluated for x = 40. To solve these problems, you need the following competences:

- Determine a profit function and a marginal profit function for given cost- and price functions.
- Differentiation in general (see learning sequences 1 and 2 of block II of our script).

### Formal approach

To solve these problems, the following four steps have to be taken consecutively (hint):

(i) To start with, determine the profit function G(x) as a function of x

(ii) Transform G(x) into the average profit function g(x)

(iii) Differentiate g(x) with respect to the quantity x

(iv) Evaluate g’(x) for x = 40.

**Step (i):** Determine the profit function G(x)
as a function of x

Profit G (G for «gain», because p might be mistaken for price) is revenue minus cost:

\(G(x) = R(x)-C(x)\).

Revenue, in turn, is price times quantity: \(R=p \cdot x\).

Thus, we start with

\(R(x)=p(x) \cdot x=(150-0.4x) \cdot x=150x-0.4x^2\)

Remark: compare this approach to *option 1* in the solution
to exercise
10 of series 5.

Since the the cost function is given – and depends on x – all we need to do is subtract C from R:

\(G(x)=R(x)-C(x)=150x-0.4x^2-(0.06x^3-x^2+50x+400)=150x-0.4x^2-0.06x^3+x^2-50x-400=-0.06x^3+0.6x^2+100x-400\)

**Step (ii):** Transform G(x) into the average
profit function g(x)

Average profit g(x) is profit per piece, so G(x) must be divided by x:

\(g(x)=\large{\frac{G(x)}{x} }\) \(= \large{\frac{-0.06x^3\,+\,0.6x^2\,+\,100x \,-\,400}{x}}\) \(=-0.06x^2+0.6x+100-400{x^{-1}}\)

**Step (iii): **Differentiate g(x) with respect
to the quantity x

Since g(x) is a polynomial function, differentiating it is easy:

\(g'(x)=-0.12x+0.6+400{x^{-2}}=-0.12x+0.6+\large{\frac{400}{x^2}}\)

**Step (iv):** Evaluate g'(x) for x = 40

In the equation g’(x), we replace x by 40 to obtain the number we have been looking for (don’t forget the units to have a correct solution):

\(g'(40)=-0.12 \cdot 40+0.6+\large{\frac{400}{40^2}}\) \(=-3.95\) (mu/qu)/qu

Remarks:

- Since the quantity units are the same in this exercise, the unit of our result is „mu“. In other exercises, this might be different!
- Economically, our result has this meaning: if, starting at a quantity of 40 units, one more unit is produced and sold, average profit goes down by about 3.95 mu.

\(G(x)=R(x)-C(x)=(150-0.4x) \cdot x-(0.06x^3-x^2+50x+400)=-0.06x^3+0.6x^2+100x-400\)

\(g(x)=\large{\frac{G(x)}{x} }\) \(= \large{\frac{-0.06x^3\,+\,0.6x^2\,+\,100x \,-\,400}{x}}\) \(=-0.06x^2+0.6x+100-400{x^{-1}}\)

\(g'(x)=-0.12x+0.6+400{x^{-2}}=-0.12x+0.6+\large{\frac{400}{x^2}}\)

\(g'(40)=-0.12 \cdot 40+0.6+\large{\frac{400}{40^2}}\) \(=-3.95\) (mu/qu)/qu