<p>Firstly, in your example the value of <span class="math-container">$r$</span> (as used by economists in this context) would be <span class="math-container">$1.03$</span>, not <span class="math-container">$0.03$</span>. Economists call this the "interest rate", but you might prefer to think of it as the "rate of return on capital".</p>

<p>Secondly, what we define as constituting one unit of capital is pretty arbitrary. Is a computer one unit of capital or ten units of capital? It doesn't make any difference, just like it doesn't matter whether you measure distance in meters or feet, so long as everyone knows what units you are using. Since this choice is arbitrary, economists usually make the simplest possible choice and measure units of capital in units of money (e.g., dollars), meaning that one unit of capital has a price of one dollar by definition: <span class="math-container">$P(K)=1$</span>.</p>

<p>If we return to your formula with <span class="math-container">$r=1.03$</span> and <span class="math-container">$P(K)=1$</span>:</p>

<p><span class="math-container">$$K[P(K)+0.03P(K)]=P(K)K(1+0.03)=Kr,$$</span></p>

<p>we get exactly the cost you saw in your course.</p>

CC BY-SA 4.0
2020-02-12T15:35:57.550Z
@Robin it's because this is a simpler expression to represent the fact that $rK$ must be greater than $K$.  Note that the expression $1.03 \times K$ can also be written $K + 0.03K$.
heh
CC BY-SA 4.0
2020-02-12T15:38:31.640Z
@Robin regarding your second question, this is almost universal convention that follows from, first of all, the role of money as a unit of account; and second of all, the fact that relative, not absolute, prices are what drives expenditure decisions.  This allows us to both "count" labor and capital using units of money, and to "normalize" that counting by dividing prices through with a convenient and otherwise arbitrary denominator.
heh
CC BY-SA 4.0
2020-02-17T10:25:49.463Z
@Robin, I beg to differ. In the given example, the value of $r$ is $0.03$, not $1.03$. See my answer below for details.
VARulle
CC BY-SA 4.0
2019-07-18T22:58:56.417Z
Thanks for the answer. Could you further explain why ‘r’ is 1.03 and not 0.03? Don’t we usually notate the interest added capital as “P(1+r)” ? (P as in Principal)
Robin
CC BY-SA 4.0
2019-07-19T06:09:28.663Z
In addition to my follow-up question, I would really appreciate it if there is an online text I can refer to that explains your answer in detail - the part where economists measure units of capital in units of money (the process of deciding one unit of capital as one dollar). I'm not saying that your answer is insufficient -  it was definitely helpful - I just wish to study this part more in detail. Thanks again for the answer!
Robin
CC BY-SA 4.0
2019-07-18T16:40:49.013Z
2019-07-18T16:40:49.013Z
Ubiquitous
<p>Just like the wage <span class="math-container">$w$</span> is the (rental) price of buying one unit (say, one person-hour) of labor you can think of <span class="math-container">$r$</span> as the (rental) price of renting one unit (say, one machine-hour of a standardized machine) of capital, such that using <span class="math-container">$K$</span> units of capital costs you <span class="math-container">$rK$</span>. In textbooks you can therefore see examples stating that e.g. the price of capital is <span class="math-container">$r$</span> = \$<span class="math-container">$10$</span>.</p>

<p>On the other hand, if the interest rate is <span class="math-container">$i=3\%=0.03$</span>, then a producer of machines in a competitive rental market would rent out a machine which costs \$<span class="math-container">$P$</span> to produce for \$<span class="math-container">$iP$</span> per year (under the usual simplifying assumptions), which translates to \$<span class="math-container">$iP/8760$</span> per hour. If you now measure capital conveniently in \$<span class="math-container">$P/8760$</span> units instead of in machine-hour units, then using <span class="math-container">$K$</span> of these (new!) units of capital costs you <span class="math-container">$iK$</span>.</p>

<p>The <span class="math-container">$r$</span> in the term <span class="math-container">$rK$</span> can therefore be interpreted both as a (hourly rental) price of capital or as a (yearly) interest rate, depending on the units in which you choose to measure capital. Since the choice of units is arbitrary, the former variant is typically used for exercises while the latter variant is used for explanations, since it establishes the macroeconomically important connection to prevailing interest rates.</p>

<p><em>Remark:</em> Using <span class="math-container">$r=1.03$</span> if the interest rate is <span class="math-container">$3\%$</span>, as the previous answer suggests, only makes sense if you assume that a machine fully depreciates within one year and has to be replaced on a yearly basis, which is not what is typically assumed.</p>

CC BY-SA 4.0
2020-02-14T11:52:09.043Z
For labour you would either rent people or buy person-hours, but you wouldn't rent person-hours.
Criticize SE actions means ban
CC BY-SA 4.0
2020-02-15T12:15:27.523Z
Thx, I corrected this.
VARulle
CC BY-SA 4.0
2020-02-12T13:07:17.200Z
2020-02-15T12:16:24.367Z
2020-02-15T12:16:24.367Z
VARulle
<p>according to the Cost formula in microeconomics class,</p>

<p>Total Fixed Cost is represented as “rK” (K as in unchanging, fixed K) </p>

<p>now my economics teacher tells me this ‘r’ is the interest rate at which you rent capital, which is pretty vague and ambiguous to me. If r represents interest “rate” and not the “interest-added cost of capital”, then how can this r be the “price” of capital?</p>

<p>So for example, if the amount of capital is 5 (K=5) I understand that you assume you rent it, or borrow the money from a bank to get the capital, incurring interest. What I don’t understand is the formula. If r is the interest rate (aka rental rate) </p>

<p>isn’t the cost formula supposed to be “K * {P(K) + P(K)*r}”, instead of “rK”?
 (P(K) here represents the price of capital per unit) </p>

<p>so if a unit of capital costs $10, and the interest rate (r) is 3%, </p>

<p>in the end you will be paying $ 5 * (10 + 10 * 0.03) in total , not 5 * 0.03 right? </p>

<p>I mean, r is just the percentge, not the ‘amount’ that you have to pay for the capital. So I don’t really understand why the textbooks and the teachers tell me that the “price” of capital is r when it seems like it’s supposed to be ‘P(K) * (1+r)’ instead of just ‘r’ . </p>

<p>I hope I can get a satisfying answer here. Would appreciate detailed explanations!  </p>

<production-function><cost><cost-functions>
Why is the price of capital ‘r’ ? (From Cost function)
Robin
# Why is the price of capital ‘r’ ? (From Cost function)

(Asked by: username_0 on 2019-07-18)

according to the Cost formula in microeconomics class,

Total Fixed Cost is represented as “rK” (K as in unchanging, fixed K)

now my economics teacher tells me this ‘r’ is the interest rate at which you rent capital, which is pretty vague and ambiguous to me. If r represents interest “rate” and not the “interest-added cost of capital”, then how can this r be the “price” of capital?

So for example, if the amount of capital is 5 (K=5) I understand that you assume you rent it, or borrow the money from a bank to get the capital, incurring interest. What I don’t understand is the formula. If r is the interest rate (aka rental rate)

isn’t the cost formula supposed to be “K \* {P(K) + P(K)\*r}”, instead of “rK”?
(P(K) here represents the price of capital per unit)

so if a unit of capital costs $10, and the interest rate (r) is 3%,

in the end you will be paying $ 5 \* (10 + 10 \* 0.03) in total , not 5 \* 0.03 right?

I mean, r is just the percentge, not the ‘amount’ that you have to pay for the capital. So I don’t really understand why the textbooks and the teachers tell me that the “price” of capital is r when it seems like it’s supposed to be ‘P(K) \* (1+r)’ instead of just ‘r’ .

I hope I can get a satisfying answer here. Would appreciate detailed explanations!

**Tags:** production-function, cost, cost-functions

---

## 2 answers

### Answer by username_1 on 2019-07-18. Score: 1 | Accepted answer

Firstly, in your example the value of $r$ (as used by economists in this context) would be $1.03$, not $0.03$. Economists call this the "interest rate", but you might prefer to think of it as the "rate of return on capital".

Secondly, what we define as constituting one unit of capital is pretty arbitrary. Is a computer one unit of capital or ten units of capital? It doesn't make any difference, just like it doesn't matter whether you measure distance in meters or feet, so long as everyone knows what units you are using. Since this choice is arbitrary, economists usually make the simplest possible choice and measure units of capital in units of money (e.g., dollars), meaning that one unit of capital has a price of one dollar by definition: $P(K)=1$.

If we return to your formula with $r=1.03$ and $P(K)=1$:

$$K[P(K)+0.03P(K)]=P(K)K(1+0.03)=Kr,$$

we get exactly the cost you saw in your course.

> **Comments:**
> **username_0**: Thanks for the answer. Could you further explain why ‘r’ is 1.03 and not 0.03? Don’t we usually notate the interest added capital as “P(1+r)” ? (P as in Principal)  
> **username_0**: In addition to my follow-up question, I would really appreciate it if there is an online text I can refer to that explains your answer in detail - the part where economists measure units of capital in units of money (the process of deciding one unit of capital as one dollar). I'm not saying that your answer is insufficient -  it was definitely helpful - I just wish to study this part more in detail. Thanks again for the answer!  
> **username_2**: @Robin it's because this is a simpler expression to represent the fact that $rK$ must be greater than $K$.  Note that the expression $1.03 \times K$ can also be written $K + 0.03K$.  
> **username_2**: @Robin regarding your second question, this is almost universal convention that follows from, first of all, the role of money as a unit of account; and second of all, the fact that relative, not absolute, prices are what drives expenditure decisions.  This allows us to both "count" labor and capital using units of money, and to "normalize" that counting by dividing prices through with a convenient and otherwise arbitrary denominator.  
> **username_3**: @Robin, I beg to differ. In the given example, the value of $r$ is $0.03$, not $1.03$. See my answer below for details.  

---

### Answer by username_3 on 2020-02-12. Score: 0

Just like the wage $w$ is the (rental) price of buying one unit (say, one person-hour) of labor you can think of $r$ as the (rental) price of renting one unit (say, one machine-hour of a standardized machine) of capital, such that using $K$ units of capital costs you $rK$. In textbooks you can therefore see examples stating that e.g. the price of capital is $r$ = \$$10$.

On the other hand, if the interest rate is $i=3\%=0.03$, then a producer of machines in a competitive rental market would rent out a machine which costs \$$P$ to produce for \$$iP$ per year (under the usual simplifying assumptions), which translates to \$$iP/8760$ per hour. If you now measure capital conveniently in \$$P/8760$ units instead of in machine-hour units, then using $K$ of these (new!) units of capital costs you $iK$.

The $r$ in the term $rK$ can therefore be interpreted both as a (hourly rental) price of capital or as a (yearly) interest rate, depending on the units in which you choose to measure capital. Since the choice of units is arbitrary, the former variant is typically used for exercises while the latter variant is used for explanations, since it establishes the macroeconomically important connection to prevailing interest rates.

*Remark:* Using $r=1.03$ if the interest rate is $3\%$, as the previous answer suggests, only makes sense if you assume that a machine fully depreciates within one year and has to be replaced on a yearly basis, which is not what is typically assumed.

> **Comments:**
> **username_4**: For labour you would either rent people or buy person-hours, but you wouldn't rent person-hours.  
> **username_3**: Thx, I corrected this.  

---

