# Multivariate Calculus

1. Mar 18, 2012

### jasper10

A firm has the following total-cost and demand functions:

C = aQ^3 - bQ^2 + cQ + d
Q = e - P

(d) Find optimizing level of Q.
(e) Chooses a,b,c,d and e such that there is only one profit-maximizing level of output Q.

I found 2 solutions for question d, but in a very long and messy form (full of the variables a-e. I do not know how to simplify my answer). Thus, I am not able to do question e). However, by analysing the marginal cost function, I found that a>0, b>0, c>0, d>0 and b^2 < 3ac.

C = aQ^3 - bQ^2 + cQ + d

MC = dC/dQ = 3aQ^2 - 2bQ + c

The coefficient of Q^2 must be positive, in order for the cost function to be U-shaped (MC must be U-shaped to make economic sense). Thus, a>0.

MC' = dMC/dQ^2 = 6aQ - 2b = 0
Hence, Q = b/3a

As Q must be positive, and a is positive, b must necessarily be positive: b>0

MCmin = 3a(b/3a)^2 - 2b(b/3a) + c
= b^2/3a - 2b^2/3a + c
= -b^2/3a + c
=(-b^2 + 3ac)/3a

thus, b^2 < 3ac and c > 0

d > 0 in order to make economic sense (it is a fixed cost).

I also found the profit function

= eQ - Q^2 - aQ^3 + bQ^2 - cQ - d

and its derivative

= e - 2Q - 3aQ^2 + 2bQ - c = 0

and solved for q

q = (-2b-2 +/- root(4b^2 - 8b + 4 + 12ae - 12ac)) / -6a

Unfortunately, from here on, I'm stuck.

2. Mar 20, 2012

### Staff: Mentor

I see one small mistake in your maths. I make it:
Code (Text):
e - 2Q - 3aQ^2 + 2bQ - c = 0

- 3aQ^2 + (2b - 2)Q + (e - c) = 0

Q = (2-2b) ± √{(2b-2)^2 -4(-3a)(e-c)} ÷ (-6a)

= (2-2b) ± √{4b^2 - 8b + 4 + 12ae - 12ac} ÷ (-6a)

= (1-b) ± √{b^2 - 2b + 1 + 3ae - 3ac} ÷ (-3a)
I can't help further, unless you are required to make Q have only one value. In which case you'd set the term under the square root to equal 0. (I know nothing about economics; I'm just guessing. So you'd need to solve b^2 - 2b + 1 + 3ae - 3ac = 0, if this were so.)