Differentiation

NZ Level 8 (NZC) Level 3 (NCEA) [In development]

Small changes and marginal rates

Lesson

An application of differentiation in the world of economics is that which surrounds marginal cost, marginal revenue and marginal profit.

All of these relate to an instantaneous rate of change (which we now know to be called the derivative) of some cost, revenue or profit functions.

Definitions

- Marginal Cost: the derivative of the cost function with respect to the production level
- Marginal Revenue: the derivative of the revenue function with respect to the production level
- Marginal Profit: the derivative of the profit function with respect to the production level

The cost, $\$C$$`C`, of producing $n$`n` coffee tables is given by

$C(n)=400n-n^2$`C`(`n`)=400`n`−`n`2 for $0\le n\le200$0≤`n`≤200

a) find the additional cost when production is increased from $15$15 to $16$16 coffee tables.

For $15$15 coffee tables the cost can be calculated using $C(15)$`C`(15)

$C(15)=400\times15-15^2=6000-225=\$5775$`C`(15)=400×15−152=6000−225=$5775

For $16$16 coffee tables the cost can be calculated using $C(16)$`C`(16)

$C(16)=400\times16-16^2=6400-256=\$6144$`C`(16)=400×16−162=6400−256=$6144

So the additional cost is $6144-5775=369$6144−5775=369

b) Calculate the marginal cost to produce $15$15 coffee tables.

This means, we need to find the derivative at the point of $n=15$`n`=15.

Derivative $C'(n)=400-2n$`C`′(`n`)=400−2`n`

$C'(15)=400-2\times15=400-30=370$`C`′(15)=400−2×15=400−30=370

The revenue function for the production of watches is given by $R(n)=n[14-(\frac{n}{1000}]$`R`(`n`)=`n`[14−(`n`1000], and the cost function for the watches is given by $C(n)=4n+7000$`C`(`n`)=4`n`+7000

a) What is the profit function?

The profit is the total revenue (in) less the total costs (out). So $P(n)=R(n)=C(n)$`P`(`n`)=`R`(`n`)=`C`(`n`)

$P(n)=n\left(14-\frac{n}{1000}\right)-\left(4n+7000\right)$`P`(`n`)=`n`(14−`n`1000)−(4`n`+7000)

$P(n)=\left(14n-\frac{n^2}{1000}\right)-\left(4n+7000\right)$`P`(`n`)=(14`n`−`n`21000)−(4`n`+7000)

$P\left(n\right)=14n-\frac{n^2}{1000}-4n-7000$`P`(`n`)=14`n`−`n`21000−4`n`−7000

$P\left(n\right)=10n-\frac{n^2}{1000}-7000$`P`(`n`)=10`n`−`n`21000−7000

b) What is the marginal profit function?

Marginal profit means to find $P'(n)$`P`′(`n`)

$P'(n)=10-\frac{n}{500}$`P`′(`n`)=10−`n`500

c) What is the marginal profit for $2000$2000 watches? Interpret this result.

This means we need to find $P'(2000)$`P`′(2000)

$P'(2000)=10-\frac{2000}{500}=10-4=6$`P`′(2000)=10−2000500=10−4=6

This means that the change in profit from making 2000 watches to 2001 watches is $6.

d) What is the optimum number of watches needed to maximise the profit?

So 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.

Set $P'(n)=0$`P`′(`n`)=0 and solve.

$P'(n)=10-\frac{n}{500}$`P`′(`n`)=10−`n`500

$10-\frac{n}{500}=0$10−`n`500=0

$10=\frac{n}{500}$10=`n`500

$5000=n$5000=`n`

So to maximise profit we should aim to manufacture $5000$5000 watches.

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+7`x`+0.002`x`2

Determine the marginal cost function.

Hence calculate the marginal cost, $C'\left(90\right)$

`C`′(90), when $90$90 items have already been produced.

Consider the function $y=$`y`=$f\left(x\right)=\frac{4}{x^2}+\sqrt{x}$`f`(`x`)=4`x`2+√`x`

Determine $f'\left(x\right)$

`f`′(`x`)Express your answer in positive index form.

Use the formula $\delta y$

`δ``y`$\approx$≈$f'\left(x\right)\times\delta x$`f`′(`x`)×`δ``x`to calculate the approximate change in $y$`y`when $x$`x`changes from $1$1 to $1.001$1.001Express your answer as an exact value.

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 $\delta y$

`δ``y`$\approx$≈$f'\left(x\right)\times\delta x$`f`′(`x`)×`δ``x`, 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.

Choose and apply a variety of differentiation, integration, and antidifferentiation techniques to functions and relations, using both analytical and numerical methods

Apply differentiation methods in solving problems