2 Applications
2.1 Graphical Analysis of Cubic
Cost Function
[see example H-L-M-R-S p.213 and 219.]
You are given the following information:
T C is total cost, T R is total revenue, ¼ are
profits, x is output, 20 is fixed cost.
T C(x) = 0:1x3 ¡ 3x2 + 50x + 20;
T R(x) = 30x
and
¼(x) = ¡0:1x3 + 3x2 ¡ 20x ¡ 20:
where lines are dashed (- - -), continuous (—)
and dotted (. . .) respectively.
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1000
800
600
400
200
0
5
10
15
20
25
x
-200
Graph of T C (- - -), T R (—–), and ¼ (. . .).
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Calculate, marginal costs, averave variable costs and marginal revenues.
MC(x) = :3x2 ¡ 6x + 50;
AV C(x) = 0:1x2 ¡ 3x + 50
and
MR = 30:
where lines are dashed (- - -), continuous (—)
and dotted (. . .) respectively.
Find the minimum of average variable
cost function.
AV C = 0:1x2 ¡ 3x + 50 Candidate(s)
for extrema: f27: 5g ; at ffx = 15:0gg
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40
30
20
10
0
5
10
15
x
20
25
30
Graph of MC (- - -), AV C (—) and MR (. . .).
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2.2 Algebraic analysis of the
cubic cost function
the maximum is obtained imposing the
equilibrium condition (first order necessary
condition)
p = MC
30 = :3x2 ¡ 6x + 50
solving ¡:3x2 + 6x ¡ 20 = 0, roots:
4: 226 5
: the former is a local minimum,
15: 774
profits are negative (check graph first and
second derivative later) the other is a local
maximum, and profits are positive.
total profits at the maximum are
¡0:1(15:8)3+3(15:8)2¡20(15:8)¡20 = 18: 489
if fixed costs are equal to 20, then there are
positive profits.
if fixed costs are equal to 40, then
¡0:1(15:8)3+3(15:8)2¡20(15:8)¡40 = ¡1: 511 2
i.e., there are some losses.
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total revenue
30 £ 15:8 = 474:0
total variable costs
0:1(15:8)3 ¡ 3(15:8)2 + 50(15:8) = 435: 51
operating surplus= total revenue-total variable
costs:
474:0 ¡ 435: 51 = 38: 49
these are used to cover fixed costs
38: 49 ¡ 40 = ¡1: 51
if operating surplus is negative, it is better to
close and just pay the fixed costs, otherwise
both fixed and variable costs are to be repaid.
average variable costs at the maximum
are
0:1(15:8)2 ¡ 3(15:8) + 50 = 27: 564
these are lower than the price, that is equal
to 30:in this case it is better to stay in the
market, at least in the short run, because
p > AV C(x):
when the AVC is at the minimum level of
15, then MC = AV C .
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MC = :3(15)2 ¡ 6(15) + 50 = 27: 5
AV C = :1(15)2 ¡ 3(15) + 50 = 27: 5
In what follows we use second derivatives. Take the profit function ¼(x) =
¡0:1x3 + 3x2 ¡ 20x ¡ 20 and differentiate to
get ¼0(x) = ¡: 3x2 + 6x ¡ 20: This expression has two solutions, 4:2 and 15:8. second
order condition ¼ 00(x) = ¡:6x + 6, so
¼00(4:2) = ¡:6(4:2) + 6 = 3:48 > 0 and
¼ 00(15:8) = ¡:6(15:8) + 6 = ¡3:48 < 0: then
we conclude the first is a local minimum and
the second is a local maximum, as the visual
inspection of the graph showed.
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