Optimizing Level of Output for Profit-Maximizing Firm

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The discussion focuses on determining the profit-maximizing level of output (Q) for a firm with specified total-cost and demand functions. The marginal cost (MC) function is derived, revealing conditions for coefficients a, b, c, and d to ensure economic viability, such as a > 0 and b^2 < 3ac. A profit function is established, and its derivative is set to zero to find Q, leading to a quadratic equation. Participants identify a mistake in the math and suggest that for a unique solution for Q, the discriminant of the quadratic should be set to zero. The conversation highlights the complexities of optimizing output in economic models.
jasper10
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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.

Any advice?
 
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jasper10 said:
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.

Any advice?
I see one small mistake in your maths. I make it:
Code:
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.)
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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