Marginal rates
An application of differentiation in the world of economics is that which surrounds marginal cost, marginal revenue and marginal profit.
Marginal cost, marginal revenue, and marginal profit all involve how much a function goes up (or down) as $x$x (the production level) increases by $1$1 unit.
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| Cost function with tangent | Approximate increase in cost using tangent |
With a cost function, $C\left(x\right)$C(x), moving along the curve cost function $1$1 unit to the right gives the exact increase in cost of producing one more item. Looking at the diagram above, the figure on the right, observe that the extra cost goes up to the blue curve, but that the marginal cost goes up a small additional amount to the green tangent line. Thus, the marginal cost is a close approximation to the cost of producing one additional item. In this case, the marginal cost will be a small amount more than the extra cost and if the cost function happened to be concave up instead of concave down like it is here, the marginal cost would be a little bit less than the extra cost.
The marginal cost described this way is the gradient of the tangent at a given level of production. Thus, we define $C'\left(x\right)$C′(x) to be the marginal cost function and evaluated at a given point $x=n$x=n it gives the approximate increase in cost of producing the $\left(n+1\right)$(n+1)th item.
Marginal revenue and marginal profit are defined similarly with the respective revenue and profit functions.
- Marginal cost function: the derivative of the cost function with respect to the production level
- Marginal revenue function: the derivative of the revenue function with respect to the production level
- Marginal profit function: the derivative of the profit function with respect to the production level
Worked example
Example 1
The revenue function for the production of watches is given by $R\left(n\right)=n\left(14-\frac{n}{1000}\right)$R(n)=n(14−n1000) and the cost function for the watches is given by $C\left(n\right)=4n+7000$C(n)=4n+7000.
(a) What is the profit function?
Think: The profit is the total revenue (in) less the total costs (out).
Do:
| $P\left(n\right)$P(n) | $=$= | $R\left(n\right)-C\left(n\right)$R(n)−C(n) |
| $=$= | $n\left(14-\frac{n}{1000}\right)-\left(4n+7000\right)$n(14−n1000)−(4n+7000) | |
| $=$= | $14n-\frac{n^2}{1000}-4n-7000$14n−n21000−4n−7000 | |
| $=$= | $10n-\frac{n^2}{1000}-7000$10n−n21000−7000 |
(b) What is the marginal profit function?
Think: Marginal profit means we need to find $P'\left(n\right)$P′(n).
Do:
$P'\left(n\right)=10-\frac{n}{500}$P′(n)=10−n500
(c) What is the marginal profit for $2000$2000 watches? Interpret this result.
Think: We need to find and interpret $P'\left(2000\right)$P′(2000).
Do:
| $P'\left(2000\right)$P′(2000) | $=$= | $10-\frac{2000}{500}$10−2000500 |
| $=$= | $10-4$10−4 | |
| $=$= | $6$6 |
This means that the approximate change in profit from making $2000$2000 watches to $2001$2001 watches is $\$6$$6.
(d) What is the optimum number of watches needed to maximise the profit?
Think: We need to maximise the profit function. The maximum profit happens when the marginal profit function is $0$0. This means that at that point, there is no extra profit to be made to make another watch.
Do: Set $P'\left(n\right)=0$P′(n)=0 and solve:
| $P'\left(n\right)$P′(n) | $=$= | $0$0 |
| $\therefore\ 10-\frac{n}{500}$∴ 10−n500 | $=$= | $0$0 |
| $\frac{n}{500}$n500 | $=$= | $10$10 |
| $n$n | $=$= | $5000$5000 |
Thus, to maximise profit we should aim to manufacture $5000$5000 watches.
Practice questions
Question 1
The cost $C$C, in dollars, of producing $x$x items of a product, is modelled by the function $C\left(x\right)=1300+7x+0.002x^2$C(x)=1300+7x+0.002x2
Determine the marginal cost function.
Hence calculate the marginal cost, $C'\left(90\right)$C′(90), when $90$90 items have already been produced.
Question 2
The revenue $R$R, in dollars, earned from selling $x$x items is modelled by the function $R\left(x\right)=-0.02x^2+40x+4000$R(x)=−0.02x2+40x+4000
Determine the marginal revenue function.
Hence calculate the marginal revenue, $R'\left(80\right)$R′(80), when $80$80 items have been sold.
Solve for the maximum number of items, $x$x, that can be sold before revenue begins to fall.
Estimating change
We can expand the concept of estimating the marginal cost to estimating the change in the dependent variable of a function. The tangent at a point gives a linear approximation to the curve at that point. As such, provided we select a value of the independent variable close to the point of contact, the tangent can calculate a reasonable approximation for the change in the value of the function. Which we can also relate as follows:
For a small change, $\delta x$δx, in the independent variable $x$x, the gradient of the secant between $A$A and $B$B will be approximately the gradient of the tangent at $A$A.
| Hence, | $\frac{\delta y}{\delta x}$δyδx | $\approx$≈ | $\frac{dy}{dx}$dydx |
| Therefore, | $\delta y$δy | $\approx$≈ | $\frac{dy}{dx}\times\delta x$dydx×δx |
Given a small change in the independent variable, $\delta x$δx, the corresponding change in the dependent variable, $\delta y$δy, can be estimated by:
$\delta y\approx\frac{dy}{dx}\times\delta x$δy≈dydx×δx
More specifically, the change in $y=f(x)$y=f(x), from $x=a$x=a to $x=a+\delta x$x=a+δx, can be estimated by:
$\delta y\approx f'\left(a\right)\times\delta x$δy≈f′(a)×δx
And the percentage change (or percentage error) can be calculated by:
| Percentage change | $=$= | $\frac{\text{Change}}{\text{Original}}\times100%$ChangeOriginal×100% |
| $=$= | $\frac{\delta y}{y}\times100%$δyy×100% | |
| $\approx$≈ | $\frac{\frac{dy}{dx}\times\delta x}{y}\times100%$dydx×δxy×100% |
Note: The marginal cost, profit and revenue calculations correspond to $\delta x=1$δx=1.
Worked example
Example 2
Consider the volume $V$V of a balloon that is in the shape of a sphere of radius $r$r cm.
(a) Determine $V'\left(r\right)$V′(r).
Think: The volume of a sphere is given by $V=\frac{4}{3}\pi r^3$V=43πr3, differentiate this in terms of $r$r.
Do:
| $V'\left(r\right)$V′(r) | $=$= | $3\times\frac{4}{3}\pi r^{3-1}$3×43πr3−1 |
Using the power rule |
| $=$= | $4\pi r^2$4πr2 |
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(b) Find the approximate change in the volume of a spherical balloon when the radius changes from $5$5 cm to $5.01$5.01 cm, using $\delta y\approx\frac{dy}{dx}\times\delta x$δy≈dydx×δx.
Think: The given formula for $\delta y$δy can be adapted for the volume function:
$\delta V\approx V'\left(r\right)\times\delta r$δV≈V′(r)×δr
We are estimating the change in $V$V from $r=5$r=5 to $r=5.01$r=5.01, for our formula we require $V'\left(5\right)$V′(5) and the change in $r$r, $\delta r=0.01$δr=0.01.
Do:
| $\delta V$δV | $\approx$≈ | $V'\left(r\right)\times\delta r$V′(r)×δr |
| $=$= | $V'\left(5\right)\times0.01$V′(5)×0.01 | |
| $=$= | $4\pi\left(5\right)^2\times0.01$4π(5)2×0.01 | |
| $=$= | $100\pi\times0.01$100π×0.01 | |
| $=$= | $\pi$π |
Thus, if the radius of the balloon changed from $5$5 cm to $5.01$5.01 cm the volume would change by approximately $\pi$π cm3.
Reflect: We can check the actual change by calculating $V\left(5.01\right)-V\left(5\right)$V(5.01)−V(5) to see if our answer is a reasonable approximation.
(c) Find the approximate percentage change in the volume of the balloon that corresponds to a $3%$3% increase in its radius, using $\delta y\approx\frac{dy}{dx}\times\delta x$δy≈dydx×δx.
Think: To get the percentage change, we need to divide the change in $V$V by $V$V, and multiply this by $100%$100%. That is:
$\text{Percentage change}=\frac{\delta V}{V}\times100%$Percentage change=δVV×100%
The given formula for $\delta y$δy can be adapted for the volume function:
$\delta V\approx V'\left(r\right)\times\delta r$δV≈V′(r)×δr
And together we have: $\text{Percentage change}\approx\frac{V'\left(r\right)\times\delta r}{V\left(r\right)}\times100%$Percentage change≈V′(r)×δrV(r)×100%, with the change in radius being $3%$3% of $r$r or $0.03r$0.03r.
Do:
| Percentage change | $=$= | $\frac{V'\left(r\right)\times\delta x}{V\left(r\right)}\times100%$V′(r)×δxV(r)×100% |
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| $=$= | $\frac{4\pi r^2\times0.03r}{\frac{4}{3}\pi r^3}\times100%$4πr2×0.03r43πr3×100% |
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| $=$= | $3\times0.03\times100%$3×0.03×100% |
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| $=$= | $9%$9% |
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Hence, a $3%$3% change in radius will lead to approximately a $9%$9% change in volume.
Practice questions
Question 3
Consider the function $y=$y=$f\left(x\right)=\frac{4}{x^2}+\sqrt{x}$f(x)=4x2+√x
Determine $f'\left(x\right)$f′(x)
Express your answer in positive index form.
Use the formula $\delta y$δy$\approx$≈$f'\left(x\right)\times\delta x$f′(x)×δx to calculate the approximate change in $y$y when $x$x changes from $1$1 to $1.001$1.001
Express your answer as an exact value.
Question 4
Consider the surface area $S$S of a spherical balloon that has a radius measuring $r$r cm.
Determine $S'\left(r\right)$S′(r)
Using the formula $\delta y$δy$\approx$≈$f'\left(x\right)\times\delta x$f′(x)×δx, form an expression for the percentage error in the surface area of a sphere that corresponds to an error of $2%$2% in the radius.
Question 5
In a particular bay, the height in metres of the tide above the mean sea level is given by $H\left(t\right)=4\sin\left(\frac{\pi\left(t-2\right)}{6}\right)$H(t)=4sin(π(t−2)6), where $t$t is the time in hours since midnight.
Determine $H'\left(t\right)$H′(t).
Use the increments formula to calculate the approximate change in height when $t$t changes from $4$4 to $4.2$4.2.
Hence, estimate the percentage change in height when $t$t changes from $4$4 to $4.2$4.2.
Round your percentage to two decimal places.