29

prices should have already been set to maximize the trade off between profit-per-sale and volume sold But profit-per-sale depends on costs, which depends on the theft numbers, so if theft increases, the equation changes.


19

What you're describing is retail shrink. It is taken into consideration when setting prices. A business will typically have consultants come in, measure their shrink to be X percent, and prices will be adjusted accordingly. Back when I worked in retail, there was a big printout in the break room informing everyone on shrink. The 5 kinds of shrink outlined on ...


10

It is the first one, $TC(0) = FC$. This is the definition. Also consider that it is not clear what is "transformed by $q$ in some way". In case of $$ \frac{5q}{q+1} + \frac{5}{q+1} $$ are the two fractions transformed by $q$, or should I just sum them up to 5? With your function, one can rearrange it to $$ TC(q) = \frac{5}{q+1} + 5 + 5q + q^2 = -\frac{5q}{...


9

Price is set by the competition In general, the prices are set by the supply and demand for the whole market. If a merchant sells the same goods for a higher price than competitors without a corresponding advantage (location, better service, convenience) then people won't buy these goods and the merchant will earn less profit as the decrease in volume will ...


7

To elaborate a bit on the answer by user 1muflon1, in economics the word "profit" is the surplus accrued to the firm after we have subtracted from revenues all compensation of production inputs, irrespective of whether these compensations have been recorded by Accounting as expenses or not. Two examples: Suppose you run your own business, and you ...


7

This is because we are talking about economic profit not accounting profit. An economic profit takes into account opportunity cost. If you are skilled programmer that can earn $\\\$100000$ per year being employed at Google then doing something else like operating your own business incurs an opportunity cost of $\\\$100000$ per year. Hence in this situation ...


7

Your reasoning seems to be this: Under the old conditions, a widget seller calculates that the profit-maximizing price of a widget is \$1. Conditions now change (e.g. because of increased theft). Under these new conditions, she should not change her selling price, because of her earlier calculation that \$1 maximizes profits. But the above reasoning is ...


4

Short answer: Yes, it is possible. Decreasing average cost implies that marginal cost is less than average cost ($MC<AC$, which can be proved by simply taking the first derivative of $C(q)/q$). With constant marginal cost, there exists a simple linear cost function $C(q)=F+a\times q$ that satisfies the constant $MC$ condition, where the constant $F$ is ...


4

Yes, if there are non-zero fixed costs, and constant marginal cost, then average cost decreases strictly monotonically with quantity, asymptotic to the marginal cost.


4

Hint For profit maximization, either $x_1$ or $x_2$ (but not both) must be zero. If not, say $x_1^*>x_2^*>0$ at the optimum, then one could increase profit by lowering cost by reducing $x_2^*$ without affecting output and thus revenue. Let $z=\max\{x_1,8x_2\}$. The profit function can be written as \begin{equation} p[3(x_3)^{1/3}(z)^{1/3}]-w_3x_3-...


4

Let $x(w, q)$ denote the solution to the cost minimization problem : \begin{eqnarray*} \min_{x} & \ w\cdot x \\ \text{s.t.} & \ \ f(x) \geq q \end{eqnarray*} where $f$ is the production function. Since $x(w, q)$ minimizes cost at $(w, q)$, following holds for all $w$ and for all $q$ : \begin{eqnarray*} w\cdot x(w, q) \leq w\cdot x(w', q) \ \ \ \...


3

Assuming a firm is a perfect competitor in input markets, the long-run average cost curve, which traces out the minimums of short-run average cost curves, can be used to characterize economies and diseconomies of scale for a firm. This is definitely a practical concept in business and IO. In an industry where the long-run average cost curve is always ...


3

You're right. Divide Eq (1) by Eq (2): $$ \frac{a L^{a-1}K^b}{bL^aK^{b-1}} = \frac{aK}{bL} = \frac{w}{r} ~~~\Rightarrow~~~ L = \frac{ar}{bw}K \tag{4} $$ Now use this in Eq. (3) $$ C = wL + rK = \left(\frac{a}{b} + 1\right)rK ~~~\Rightarrow~~~ K = \frac{C}{r(a/b + 1)} \tag{5} $$ Replace this in Eq. (4) to get $L$


2

I understand that the "bundle price" is cost to us. Then your table depicts a linear system of equations $$C_j = a_1x_1+a_2x_2+...+a_mx_m,\;\;\; j=1,...,n$$ with $C_j$ and the $x$'s being the known quantities, and we want to determine the unknown alphas. If it so happens that the bundles on offer are same in number as the $x$'s, ($n=m$), then the system,...


2

When you say non-linear cost function, I assume you aren't referring to the firm's production having non-linear costs, but judging by your example, you rather mean the firm's output prices having weird optimal bundle pricing. From the abstract of the linked article (Hanson and Martin, 1990): " Bundle pricing is a widespread phenomenon. However, despite ...


2

So to understand why the long run average cost curve and short run average cost curve have the same minimum in perfect competition, as well as some of the other stuff you ask, you have to understand the different assumptions that underlie the models you're working with. Let's start with what the short run and long run mean here. In this context, the short ...


2

If your cost function is also homogeneous of degree $k$ (which is often assumed to model different types of returns to scale, whether constant, increasing, or decreasing), then by Euler's Homogeneous Function Theorem, $$ x c'(x) = k c(x).$$ That is, $x c'(x)$ is your cost itself, up to some scaling factor $k$ (for example, if $c(x) = ax$ so that $c(x)$ is ...


2

In general the statement is wrong. Here is a counterexample: Suppose you have $f(k,l) = -k l^\beta$ with $\beta >0$ and $(k,l)\in\mathbb{R}^2_{++}$ (you can interpret $f$ as a production function for a "bad" commodity). Then you have: $$f(tk,tl) = - t^{1+\beta} k l^\beta = t^{1+\beta} f(k,l) < t f(k,l)$$ so that $f(k,l)$ is decreasing returns to scale....


1

Investopedia has this definition: Economic profit = revenues - explicit costs - opportunity costs where opportunity costs must be either zero or positive based on management evaluation of at least one counter-factual scenario. This article argues that opportunity costs are zero or positive and cannot be negative because logically a negative cost would be a ...


1

I see your point. However, theft in effect increases the cost of purchase, which changes everything. The easiest way to get to the bottom of the answer is thinking of a simple scenario but in extremes. A shop sells one item only, has no thefts, purchases the items for 10 each, and sells them for 15 (making 5), and sells 10 per hour (making 50 profit per hour)...


1

Profits of a price taking firm take the form $$ p q - c,$$ where $p$ is the price, $q$ is the output and $c$ is the cost. Of course there is a relation between the output $q$ and the cost $c$ in the sense that higher $q$ will correspond to higher $c$. As such, these two quantities cannot be chosen independently. There are two possibilities to make this ...


1

Either I totally misunderstand the question, or this is just adding up numbers, there are no economic or mathematical principles involved. (In which case this is not the right SE for it.) Denoting the "balances" by $b_1,b_2,...$ and the "supplies" by $s_1,s_2,...$ total profits are $$ \Pi_{total} = \sum_i (b_i-s_i) $$ You want everyone to end up with a ...


1

Look at it this way : There are two timeframes in which firms make decisions - Short Run and Long Run. The actual time that these "Runs" include vary from Industry to Industry. For example, for a wheat grower, a full one year can be seen as long-run whereas, for a thermal power plant, 15 years would imply the long run. The answer to your question lies in ...


1

The $r$ in the total cost function is the interest rate, which at the same time constitutes the price of capital. You can think of it as an explicit cost, e.g. if you rent the capital or pay interest to the bank, or as an implicit (opportunity) cost, e.g. if you own the capital and by using it yourself you give up the opportunity to rent it out.


1

I'm just gonna try doing it. The problem is: $$\text{min } C = rK + wL \\ \text{s.t. } q = K + \ln(L) $$ The implied constraint, of course, is $L \in [1, \infty)$ (non-negativity constraint, as $q > 0$, and $K = 0$ is, of course, possible). We can try solving this using Lagrange multipliers. Define the Lagrangian: $$\mathcal{L} = (rK + wL) + \lambda(q ...


1

I`d propose you to follow these steps: Set up the minimization cost problem (i.e. for a given output quantity $y$ minimize costs): \begin{align} \min_{H,L,K}& \quad sH + wL + rK \tag{1} \label{1}\\ \text{such that} &\quad \min\{H,L\} + \min\{H, K\}\geq y \tag{2} \label{2} \end{align} In principle you have 3 cases, depending on price of factors $(s,...


1

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 ...


1

Max Then you have: Max Which gives you: So L = 100, and you have a loss of 15000


1

There is no deeper meaning than what you already discuss. The Total Cost function is always made up of fixed costs plus the variable costs (each of which may be zero). We have: $TC(Q) = F \; + \, VC(Q)$ where $TC$ is the total cost function, $F$ are the fixed costs and $VC$ are the variable costs. Fixed costs do not depend on the quantity, unlike ...


1

We examine the function $F(K,L)$ that is homogeneous of degree $\lambda < 1$. Then we have that its partial derivatives are homogeneous of degree $\lambda -1 $. For a homogeneous function $F(K,L)$ of degree $\lambda$ it holds that $$K\cdot F_K + L\cdot F_L = \lambda \cdot F(K,L) \tag{1}$$ Analogously for the partial derivatives we have $$F_L:\;K\cdot ...


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