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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 $$TC(q) = \frac{5}{q+1} + 5 + 5q + q^2 = -\frac{5q}{... 4 Short answer: Yes, it is possible. Decreasing average cost implies that marginal cost is less than average cost (MC<AC, which can be proved by simply taking the first derivative of C(q)/q). With constant marginal cost, there exists a simple linear cost function C(q)=F+a\times q that satisfies the constant MC condition, where the constant F is ... 4 Yes, if there are non-zero fixed costs, and constant marginal cost, then average cost decreases strictly monotonically with quantity, asymptotic to the marginal cost. 4 Hint For profit maximization, either x_1 or x_2 (but not both) must be zero. If not, say x_1^*>x_2^*>0 at the optimum, then one could increase profit by lowering cost by reducing x_2^* without affecting output and thus revenue. Let z=\max\{x_1,8x_2\}. The profit function can be written as \begin{equation} p[3(x_3)^{1/3}(z)^{1/3}]-w_3x_3-... 4 Let x(w, q) denote the solution to the cost minimization problem : \begin{eqnarray*} \min_{x} & \ w\cdot x \\ \text{s.t.} & \ \ f(x) \geq q \end{eqnarray*} where f is the production function. Since x(w, q) minimizes cost at (w, q), following holds for all w and for all q : \begin{eqnarray*} w\cdot x(w, q) \leq w\cdot x(w', q) \ \ \ \... 3 Assuming a firm is a perfect competitor in input markets, the long-run average cost curve, which traces out the minimums of short-run average cost curves, can be used to characterize economies and diseconomies of scale for a firm. This is definitely a practical concept in business and IO. In an industry where the long-run average cost curve is always ... 3 You're right. Divide Eq (1) by Eq (2):$$ \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} $$Replace this in Eq. (4) to get L 2 In general the statement is wrong. Here is a counterexample: Suppose you have f(k,l) = -k l^\beta with \beta >0 and (k,l)\in\mathbb{R}^2_{++} (you can interpret f as a production function for a "bad" commodity). Then you have:$$f(tk,tl) = - t^{1+\beta} k l^\beta = t^{1+\beta} f(k,l) < t f(k,l)$$so that f(k,l) is decreasing returns to scale.... 2 I understand that the "bundle price" is cost to us. Then your table depicts a linear system of equations$$C_j = a_1x_1+a_2x_2+...+a_mx_m,\;\;\; j=1,...,nwith 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 ... 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`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 thatK\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 ...
Short answer: whether a firm's maximum 'profit' at the point where marginal revenue equals marginal cost is positive or negative depends on whether average revenue exceeds average cost at that point. To give a simple numerical example, suppose for a firm in a perfectly competitive market: $$MR = AR = 1$$ $$TC = 2 + 0.4Q + 0.1Q^2$$ where $MR$ is marginal ...