## Some Applications of Derivatives — Part II

Derivatives in Economics.

Engineers use the terms velocity and acceleration to refer to the derivatives of functions describing motion. Economists, too, have a specialized vocabulary for rates of change and derivatives. They call them marginals.

In a manufacturing operation, the cost of production c(x) is a function of x, the number of units produced. The marginal cost of production is the rate of change of cost (c) with respect to a level of production (x), so it is $dc/dx$.

For example, let c(x) represent the dollars needed needed to produce x tons of steel in one week. It costs more to produce x+h units, and the cost difference, divided by h, is the average increase in cost per ton per week:

$\frac{c(x+h)-c(x)}{h}=$ average increase in cost/ton/wk to produce the next h tons of steel

The limit of this ratio as $h \rightarrow 0$ is the marginal cost of producing more steel when the current production level is x tons.

$\frac{dc}{dx}=\lim_{h \rightarrow 0} \frac{c(x+h)-c(x)}{h}=$ marginal cost of production

Sometimes, the marginal cost of production is loosely defined to be the extra cost of producing one unit:

$\frac{\triangle {c}}{\triangle {x}}=\frac{c(x+1)-c(x)}{1}$

which is approximately the value of $dc/dx$ at x. To see why this is an acceptable approximation, observe that if the slope  of c does not change quickly near x, then the difference quotient will be close to its limit, the derivative $dc/dx$, even if $\triangle {x}=1$. In practice, the approximation works best for large values of x.

Example: Marginal Cost

Suppose it costs $c(x)=x^{3}-6x^{2}+15x$  dollars to produce x radiators when 8 to 30 radiators are produced. Your shop currently produces 10 radiators a day. About how much extra cost will it cost to produce one more radiator a day?

Example : Marginal tax rate

To get some feel for the language of marginal rates, consider marginal tax rates. If your marginal income tax rate is 28% and your income increases by USD 1000, you can expect to have to pay an extra USD 280 in income taxes. This does not mean that you pay $28$ percent of your entire income in taxes. It just means that at your current income level I, the rate of increase of taxes I with respect to income is $dT/dI = 0.28$. You will pay USD 0.28 out of every extra dollar you earn in taxes. Of course, if you earn a lot more, you may land in a higher tax bracket and your marginal rate will increase.

Example: Marginal revenue:

If $r(x) = x^{3}-3x^{2}+12x$ gives the dollar revenue from selling x thousand candy bars, $5<= x<=20$, the marginal revenue when x thousand are sold is

$r^{'}(x) = \frac{d}{dx}(x^{3}-3x^{2}+12x)=3x^{2}-6x+12$.

As with marginal cost, the marginal revenue function estimates the increase in revenue that will result from selling one additional unit. If you currently sell 10 thousand candy bars a week, you can expect your revenue to increase by about $r^{'}(10) = 3(100) -6(10) +12=252$ USD, if you increase sales to 11 thousand bars a week.

Choosing functions to illustrate economics.

In case, you are wondering why economists use polynomials of low degree to illustrate complicated phenomena like cost and revenue, here is the rationale: while formulae for real phenomena are rarely available in any given instance, the theory of  economics can still provide valuable guidance. the functions about which theory speaks can often be illustrated with low degree polynomials on relevant intervals. Cubic polynomials provide a good balance between being easy to work with and being complicated enough to illustrate important points.

Ref: Calculus and Analytic Geometry by G B Thomas.

More later,

Nalin Pithwa

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