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"I am not given wealth $w$ although I suppose I could assume any firm who is purchasing has some budget." No. This is exactly where the fundamental microeconomic theory of the firm differs from the microeconomic Consumer Theory: the firm is not constrained by a budget. The reason is that this fundamental theory deals most and foremost with the "long-term" ...

9

What he's saying is that once the burger is already made, the cost of making the burger is a sunk cost, and thus the marginal cost of the burger is just the cost of the tiny labor involved in picking it up and selling it to the customer. That's sort of an odd position to take. Obviously once the burger is cooked you can't recuperate the costs, but in the ...

8

This question really forces one to think about the role that quantity plays in the competitive equilibrium. The two main points that, I think, explain the way this works are: The market quantity is endogenous In competitive equilibrium, the market clears I think the thing that is perhaps causing confusion here is that, recalling that it is a true statement ...

7

As said in my comment, The fixed cost is usually defined as the cost when quantity is equal to zero, and the variable cost as the total cost minus the fixed cost. Hence, if $TC(q)$ is the total cost for the given level of quantity $q$, then $FC=TC(0)$ is the fixed cost, which is a constant independent of $q$; and $VC(q)=TC(q)-FC$ is the variable cost. ...

6

Your are right. You have to minimize the average cost. $$c(Q)=\frac{C(Q)}{Q}=6000 +40Q+Q^2$$ Calculate the first derivative and set it equal to zero: $c'(Q)=40+2Q=0$ Solve this equation for $Q$. Denote the optimal value as $Q^*$. $Q^*$ can be a local maximum or a local minimum If $c''(Q^*)>0$, then you have found the local minimum. The local ...

5

The derivative is used in some contexts, but not all, when the cost function is differentiable. In those contexts, it tends to be assumed that supply is continuous, not discrete. This is a matter of convention and of analytic convenience. It has the advantage of being consistent, whether you're approaching the supply point from above or from below. But in ...

5

You might want to read up on repeated games. You are right, in a one-period model, once produced, the seller has little marginal cost, so could potentially sell at any price. However, his price at $t$ will affect behavior at $t+1$. He needs to credibly commit (or signal) that he will not do this again at $t+1$, otherwise he will be stuck in the same ...

5

The people who do the planning do explicitly account for the value of time and convenience to transit users. At least, in all the systems I've worked on. (source: I have had jobs where I worked on exactly this, for London, for the Netherlands, and for international train travel) Interchanges are modelled with penalties which cover: the time taken to move ...

4

Since it was mentioned in an another answer let's clear this first: whether the transportation (and its time and monetary costs) should be associated with the intended consumption of the good you are going to purchase, or it can be considered as consumption on its own, depends on your subjective view of it: do you derive any form of pleasure by the trip ...

4

Suppose initially there are no fixed costs. What does it mean to take an average? Consider a cost function $C(y)$. What does it mean to take the “average” of this function? Mathematically, it is just $$A(y) = \frac{c(y)}{y}$$ Let’s suppose we are considering $C(y) = y^3$. Suppose we now consider $y = 5$. Then $$A(5) = \frac{5^3}{5} = 25$$ This is just ...

4

You're making this way more complicated than it needs to be. Edit: Okay it's a little more complicated that I thought but hey! What a cool result! $AC = \frac{C(q)}{q} \\ MC = C'(q)$ When you minimize $AC$ with respect to $q$, $$\frac{\partial AC}{\partial q} = \frac{C'(q) \cdot q - C(q)}{q^2} = 0$$ $$\implies C'(q) \cdot q - C(q) = 0$$ $$\implies C'(q) \... 4 Yes, if there are non-zero fixed costs, and constant marginal cost, then average cost decreases strictly monotonically with quantity, asymptotic to the marginal cost. 4 Short answer: Yes, it is possible. Decreasing average cost implies that marginal cost is less than average cost (MC<AC, which can be proved by simply taking the first derivative of C(q)/q). With constant marginal cost, there exists a simple linear cost function C(q)=F+a\times q that satisfies the constant MC condition, where the constant F is ... 4 You are missing the average cost curve in the same diagram. Basic algebra gives us the following. Let's find the minimum of the AC = C/Q. We have$$\frac {\partial AC}{\partial Q} = \frac {MC\cdot Q - C}{Q^2}$$For this to be equal to zero, we must have MC \cdot Q = C \implies MC = AC. So when AC is at its minimum, it equals MC. But we also ... 4 I'll offer a less algebraic alternative to Alecos's answer. In short, yes and no. The "no" part Normally the MC and AC curves would look like the following, with MC intersecting AC from below AC's minimum point. Suppose price P_0 were below this point. Then the firm would sell at a quantity below Q_1. But what does this imply for the firm's profit? On ... 4 Hint For profit maximization, either x_1 or x_2 (but not both) must be zero. If not, say x_1^*>x_2^*>0 at the optimum, then one could increase profit by lowering cost by reducing x_2^* without affecting output and thus revenue. Let z=\max\{x_1,8x_2\}. The profit function can be written as p[3(x_3)^{1/3}(z)^{1/3}]-w_3x_3-... 3 To help you discern the two, let's try to explain with words and understand what information are we getting from the derivative and from the difference, respectively: The derivative gives you information about the change in cost relative to change in quantity produced, in a specific, local, point(quantity)1. In other words you are measuring the change in ... 3 The total cost was 291 m.u. for 1 unit given MC(q) \equiv \frac{dC(q)}{q} = 3 q^2 - 40 q + 220. First, the area under the Marginal Cost gives the Total Variable Cost, TVC(q). To find this area, we will integrate the Marginal Cost:$$ \begin{align} TVC(q) &= \int_0^q\frac{dTVC(q)}{dq} dq = \int_0^q (3q^2 - 40 q + 220) dq\\ &= \left. \left(q^3 -...

3

Marginal cost is always the same as the shadow price in the cost minimization problem \begin{eqnarray*} \min_x && w \cdot x \\ s.t. && f(x) = y. \end{eqnarray*} In optimum the shadow price (Lagrange multiplier) belonging to the condition $f(x) = y$ is the marginal cost. However there are other optimization problems where the shadow price is ...

3


3

Yes, there's a cost to running the New York Stock Exchange. There's the premises, utility bills, staff, systems, and so on. Yes, this affects the price of stocks on the exchange. The exchange recoups its fees, ultimately, from the companies listed on it, pushing up their costs. However, for pretty much all companies listed on it, the stock exchange fees ...

3

The marginal cost is 3. Marginal costs do not depend on the fixed cost, and when your variable costs are constant, then the marginal cost and the variable cost are the same. Note that your total cost is $C=FC+3q$ and the marginal cost is always the derivative of your total cost, in this case, $3$. As for the fixed costs, 4000 is definitely part of it, but ...

2

1) "To first order" appears to be a shorthand for "to a first-order approximation", meaning in turn, "to a linear approximation". 2) It is not that "the one implies the other", it is that the one is consistent with the other. In attempting to arrive at a functional specification for the "tax distortion costs" function, say $C_D=h(T)$, first we reason ...

2

To make it simple: let say your coconuts production is constant over time but the population is growing. It's a constant supply but an increasing demand, resulting in an increase of coconuts price. If the prices were constant, the demand would be higher and thus not fully satisfied.

2

You don't need to have any knowledge of "wealth" here, because this "wealth" that is necessary to produce is exactly the value of the cost function, it is what you are trying to find. Cost function is the minimal amount of expenditures necessary to produce a given amount of product given some prices. Suppose your the price of your final good is $p$, and ...

2

That comment said, I can provide an answer here. Take an average cost function: $$f(x) = mx + b$$ where x is output, b is the intercept, and m is the gradient. So total cost is: $$C(x) = x(mx + b) = mx^2 + bx$$ And that makes marginal cost the derivative of this function: $$C'(x) = 2mx + b$$ Best of luck with your homework.

2

The function $C(q) = q^2$ is non-linear, therefore the rate of change of $C(q)$ with respect to q is constantly changing. When you take $\frac{C(3)-C(2)} {3-2}$ you are finding the rate of change over a range of $q$, not the rate of change at $q = 3$. This is where taking a derivative is needed, because it gives you the rate of change at the point $(q,C)$ ...

2

Does average cost necessarily equal marginal cost at some point? Yes, if the Average Cost function has an interior stationary point. Considering the derivative of the Average Cost function with respect to quantity, we have $$\frac {\partial AC}{\partial y} = \frac {\partial (C/y)}{\partial y} =\frac {MC\cdot y - C}{y^2} = \frac {MC - AC}{y}$$ Then \...

2

The classic reference on this topic (as suggested by my professor) is the book; Theory of Cost and Production Functions by Ronald Shepherd.

2

MP = Change in output / Change in input For the first product MP1 = 1 / 500 For the second MP2 = 1 / 600 Hence, the MP is decreasing. Hope this helps.

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