# Marginal analysis problem

1. Jun 4, 2004

### Zorodius

This is a problem from a section on marginal analysis in a calculus textbook I have been studying. I can't ask the instructor, since I don't have one, I'm just studying this for my benefit.

The problem is as follows:

A firm's costs are given by:

$$C(q) = Kq^{1/a} + F$$

Where a is a positive constant, F is the fixed cost, and K is the technology available to the firm.

Part A: Show that C is concave down if a > 1.

Part B: Assuming that a < 1 and that average cost is minimized when average cost equals marginal cost, find what value of q minimizes the average cost.

I have no questions about part A (I take the second derivative of C with respect to q, and the result seems to make it obvious that the curve would open downward.) My questions are about part B. The book's answer is:

q = [Fa/(K(1-a))]^a

I don't understand how they arrived at that answer. Moreover, I don't understand how it would be physically possible to have F as a factor in the answer, since I was under the impression that average cost was something along the lines of C(q + 1) - C(q), which would cause the F's to cancel out, and that taking the derivative of C(q) would also cause the F's to disappear, since the derivative of a constant is zero.

I tried working backwards from their answer, and got to:

$$\frac{K}{a}q^{1/a} = F + Kq^{1/a}$$
The right hand side is the original function, and the left hand side is what the derivative of C(q) would be if you forgot to subtract one from the exponent when using the power rule.

Could someone explain how to solve this problem?

2. Jun 4, 2004

### AKG

I can't see how it is that by working backwards, you got what you got:

$$q = {\left [ \frac {Fa}{K(1 - a)} \right ]}^a$$
$$\frac{K}{a}q^{\frac{1}{a}} = \frac{F}{1-a}$$

Now, I would think average cost would be found by integrating the function over some interval, and dividing by the length of the interval, giving you the average over that interval, but there is no given interval, so I might assume that average cost simply refers to C(q). Marginal cost I would assume refers to C'(q). So, it looks like you're trying to find $q_{min}$ satisfying the equation:

$$C(q_{min}) = C'(q_{min})$$
$$K{(q_{min})}^{\frac{1}{a}} + F = \frac{K}{a}{(q_{min})}^{\frac{1}{a} - 1}$$

I can tell you, I can't see how I'd go about isolating q.

3. Jun 4, 2004

### Zorodius

$$q = [\frac {Fa} {K(1-a)}]^a$$
$$q^{1/a} = \frac {Fa} {K(1-a)}$$
$$q^{1/a} - aq^{1/a} = \frac {Fa}{K}$$
$$q^{1/a} = \frac{Fa}{K} + aq^{1/a}$$
$$Kq^{1/a} = Fa + Kaq^{1/a}$$

thus,

$$\frac {K}{a}q^{1/a} = F + Kq^{1/a}$$

I agree, but this section is prior to the first section on integration, and they use no integrals in the material up to this point, so I assume that average cost either refers to C(Q), as you said, or to some variation on that.

Me either :yuck:

4. Jun 5, 2004

### Zorodius

Anyone? Did they just screw up this question, or is there a way of correctly solving it to arrive at their answer?