9. Elasticity and economic quantities - Mathematics for Business

9.1Total, average and marginal functions¶

In this section, we are going to use the following notation:

quantity of production $\dots Q$
price $\dots p$
total cost function $\dots C(Q)$
total revenue function $\dots R(Q)$
profit function $\dots \Pi(Q) = R(Q) - C(Q)$
marginal cost function $\dots MC(Q) = C'(Q)$
marginal revenue function $\dots MR(Q) = R'(Q)$
average cost function $\displaystyle \dots AC(Q) = \frac{C(Q)}{Q}$
average revenue function $\displaystyle \dots AR(Q) = \frac{R(Q)}{Q}$
demand function $\dots q(p)$

Since the profit is equal to the difference of the revenue and the cost, we have

\Pi(x) = e^{-x+13} \cdot \left( (x-14)^2+1\right) + \frac{x^3}{3} + 225x - 15x^2.

In order to find the minimum, first we have to compute the derivative of the profit function:

\begin{split} \Pi'(x) &= (e^{-x+13})' \cdot \left((x-14)^2+1\right) + e^{-x+13} \cdot \left( (x-14)^2+1\right)' + x^2 + 225 -30x \\ &= -e^{-x+13} \cdot \left( (x-14)^2+1\right) + e^{-x+13} \cdot(2(x-14)) + x^2-30x + 225 \\ &= -e^{-x+13} \cdot (\left (x-14)^2 -2(x-14) +1 \right) + x^2 -2 \cdot 15 \cdot x + 15^2 \\ &= -e^{-x+13} \cdot \left( x-15\right)^2 + (x-15)^2 \\ &= (x-15)^2 \cdot (-e^{-x+13}+1). \end{split}

Now, the stationary points are given by:

\begin{split} \Pi'(x) &= 0 \\ \implies (x-15)^2 = 0 &\text{ or } -e^{-x+13}+1 = 0 \\ \implies x = 15 &\text{ or } x = 13. \end{split}

In order to determine which point is a local minimum, we need to compute the second derivative of the function $\Pi$ :

\begin{split} \Pi''(x) &= \left( (x-15)^2 \cdot (-e^{-x+13}+1)\right)' \\ &= 2(x-15) \cdot (-e^{-x+13}+1) + (x-15)^2 \cdot e^{-x+13}. \end{split}

Plugging-in the stationary point $x = 13,$ we get that:

\Pi''(13) = 4 > 0,

so the point $x = 13$ is a local minimum of the profit function.

9.2Elasticity¶

Interpretation of the coefficient of elasticity is that if the independent variable $x$ increases by $1\%,$ then the value of $y$ will change by approximately $E_{y,x} \%.$

Solution to Problem 9.8

So, if we want the function $f$ to be inelastic at the level $x = 1,$ then we need to compute the coefficient of elasticity, plug-in $x = 1$ and find the the absolute value of the expression we got is less than 1.

First, let’s compute the coefficient of elasticity:

\begin{split} E_{f,x} &= \frac{x}{f} \cdot f' \\ &= \frac{x}{xe^{(t+1)x}} \cdot (x \cdot e^{(t+1)x})' \\ &= \frac{1}{e^{(t+1)x}} \cdot (e^{(t+1)x} + x \cdot e^{(t+1)x} \cdot (t+1)) \\ &= \frac{1}{e^{(t+1)x}} \cdot e^{(t+1)x} \cdot (1+x(t+1)) \\ &= 1+x(t+1). \end{split}

At the level $x = 1,$ we have $E_{f,x}(1) = t+2.$

Now, we want the absolute value of the coefficient of elasticity to be less than 1, so:

\begin{split} \lvert E_{f,x}(1) \rvert &< 1 \\ \implies \lvert t+2 \rvert &< 1 \\ \implies -1 &< t+2 < 1 \\ \implies -3 &< t < -1. \end{split}

So, in order for the function $f$ to be inelastic at the level $x = 1,$ the parameter $t$ has to belong to the interval $\langle -3, -1 \rangle.$

Part 2. Functions of one variable

8. Optimization of functions of one variable

Part 3. Functions of several variables

10. Partial derivatives