Exercises 139
(b) Marginal cost is the derivative of the cost function:
dC(y)
MC(y) = = 3(y − 10)2 .
dy
To find the minimum of the marginal cost curve, we differentiate MC(y) and set the
result equal to zero:
dMC(y)
= 6(y − 10) = 0.
dy
This gives y = 10.
Average cost is
C(y) 1,000 + (y − 10)3
AC(y) = = .
y y
Differentiating AC(y) and setting the result equal to zero gives
dAC(y) y(3(y − 10)2 ) − (1,000 + (y − 10)3 )1
= = 0.
dy y2
PMG
This leads to
3y(y − 10)2 = 1,000 + (y − 10)3 ,
which simplifies to
(2y + 10)(y − 10)2 = 1,000,
which yields y = 15. Consequently, average cost is minimized at y = 15.
(c) At y = 15, the point at which average cost is minimized, average cost is
1,000 + (y − 10)3 1,000 + 125
AC(15) = = = 75.
y 15
Therefore, for any price p < 75 and any y > 0, price is less than average cost. If
p < 75, the firm will produce y = 0. For p ≥ 75, the firm will maximize profit by
setting price equal to marginal cost, which gives
p = MC(y) = 3(y − 10)2 .
√
Solving this equation for y as a function of p leads to y(p) = 10 + p/3.
Exercises
1. Let the production function be y = x 1/2 . yt
=
-(a) Show that the production function y(x) is concave.
-
(b) Suppose the price of x is w = 1. Find the firm’s total cost curve C(y), average cost
curve AC(y), and marginal cost curve MC(y).
-(c) Find the firm’s supply curve y ∗ (p).
-(d) Suppose the price of y is p = 10. Calculate the firm’s profit.
140 Chapter 8. Theory of the Firm 1: The Single-Input Model
2. Assume the production function is y = 5x 1/3 − 30, and the price of x is w = 1.
-
(a) Derive the firm’s total cost curve C(y), average cost curve AC(y), and marginal
cost curve MC(y).
-(b) What is the firm’s supply curve y ∗ (p)?
3. Consider the production function from question 1, y = x 1/2 . Assume x ≥ 1.
v
W(a) Show that the inverse production function x(y) is convex.
(b)
-The price of y is p = 10. Find the firm’s marginal product MP (x) and average
f(u) y x
=
=
product AP (x).
(c)
2Find the firm’s value of marginal product V MP (x) and value of average product
Inverse Prod" In V AP (x).
(d) Find the firm’s input demand curve x ∗ (w).
I" (8) x=
~yz
(e) Suppose the price of x is w = 1. Calculate the firm’s profit.
=
4. Suppose the production function is y = x 2/3 + 13 x, and the price of y is p = 6. Assume
x ≥ 1.
(a) Find the firm’s marginal product MP (x) and average product AP (x).
(b) Derive the firm’s value of marginal product V MP (x) and value of average product
V AP (x).
PMG
(c) What is the firm’s input demand curve x ∗ (w)?
5. Consider the single-input/multiple-output model. Recall that x = f −1 (y1 , y2 ), the
inverse production function, represents the firm’s technological constraint. Can you
solve the profit maximization problem for this firm by focusing on the input variable?
Hint: Do it with the following four steps. (Note: Because we have not specified the
f −1 function, this is a graphical exercise, without specific functional or numerical
solutions.)
(a) An isofactor curve is a locus of output combinations that use the same level of
input. In a graph of the (y1 , y2 )-quadrant, sketch some isofactor curves, assuming
f −1 is convex.
(b) An isorevenue line is a locus of output combinations that yield the same total
revenue. Plot several isorevenue lines on the same graph as the isofactor curves.
(c) Solve the revenue maximization problem for a fixed level of input. This will yield
the conditional output supply curves y1 (p1 , p2 , x) and y2 (p1 , p2 , x).
(d) Finally, write down the profit maximization problem, making profit a function of
dup the single variable x.
I 6. The inverse production function with one input and two outputs is x = y12 + y22 + y1 y2 .
Assume the price of x is w = 1.
(a) Find the firm’s total cost curve C(y1 , y2 ) and marginal cost curves MC1 (y1 ) and
X MC (y ).
2 2
(b) Find the firm’s supply curves y1∗ (p1 , p2 ) and y2∗ (p1 , p2 ), subject to the nonnegative
* profit condition.
*(c) Suppose p1 = 1 and p2 = 1. Calculate y1∗ and y2∗ . What is the firm’s profit?
(d) Suppose p2 rises to 2. Recalculate y1∗ and y2∗ . How has the firm’s profit changed?
si). (a) TC w.x
=
Mxc,(71) 8y,(8, 2, 2.32]
=
+
+
TC (I
=
u
2y, yz
+
(4,, f2) y, yc +
=
TC =
+
y, yn
Mx x(82)
6y.(2, 8 8,25]
+ +
2yz
= Y1
+
#(a) =
y
Digf w.r.t. U
I
-
:It: Iu
Diff. w.r.t.u
again
PMG
-
- 1
:. N
.
-3
:-
In CO
As
O :.y(u) is concave
fu.
t
(b) TC
(y) xw =
y
=
As w I
=
=x
y
TC(y) x
=
TC(y) y2 1C(81
II y
4y
Mc(y)= =
=
=
=
=
2y
=
(C) Supply cure
P MC=
P
2y
=
y =
PMG
(d) T TR-TC =
P M2
=
P. yz 10
it
Y 2y
=
- =
:10 x 15) -
(5)2 y
5
=
=50 -
25
25
=
Production Function Lost Function
I
q b(l)
=
Wage Rate, w TC (9):lw
put y f(x)
=
W
TC(y) x.w
=
9 f(1,k)
:
PMG
Wage Rate,
Rental Value,
w
8
-C (9):Nl+UK
M2
Cost $Rey Fiem
stry -
MS
22
P2 Pz 0
MR AR2 d,
=
=
Pi - MR, AR, =d, =
PMG
MD
MD
0
9,92 Output
P MC
=
MR
and
slope of MC>
Slope of
section AVC came
Rising of MC, which lies above is
supply
of time in
Peng Comp. In Stost-Run.
Profits, it: TR-TC
Diff. w.r.t.
9
** q(TR)
=
q
-
0 MR
=
- MC
MR MC
PMG
=
P M2
=
=
MC
MR.
Diff again w.r.t.9
dit: Slope MR-slope of inc
profit
O
For
Man,
i.e.
Slope of MCs
slope of MB
S01 6(b) P, MC, (81)
=
Pc Mcc(y)
=
+
-
D
1, 20
=
y2
+
-
y. -
Solving &
P, 2(i -2y]
=
y2
+
in
22]
=
+ 3,
2 - ie
PMG
1, 21 =
-
Yy yz
+
P, 212 = -
3yz
3y2 21 = -
41
2=y
ya:
Si
(C) P, 1
=
12 1
=
i TR -TC
=
y =
I
(,2,
- +
xyz] -
(, +
y, y,7)
+
y1
=
I (5 5) +4 5]
=
- +
- I
-
=
&(a) y=
5xs-30 W 1
=
30 5x"
y
+ =
TC a.W
=
30 as I =
(2)()
(530) Tc(y) (y)
=
PMG
a =
A((y)
3) M((y)
(y)+
= =
(y 30)
=
+
=
3(y 30)2
+
125y 125
(b) P MC =
P
55(y
=
30)
+
-P (y 30
=
+
1P y+
=
30
y P
=
- 30
Sincesse God" x=
(a)
-
x
y
=
fu] y =
=
x
2y
a
"
2
=
> 0 So, Inverse Prod" is conver
(b) MP(u), AP(n) 900 TP
L
PMG
A
MP(n):8 Me:
=
(x) Al:
I
Mp(n):
ete l
AP1) =
==Fu
(C) XMP = P. MP XAP PXAR
=
10 I
10 x
m
-
2.T ·
-Fr :
Method
-
-
- I
Md-2
(d) XMP = W
i TR
=
- TC
Ec w
=
= P.
y-W.x
- I 10.-W.x
=
u
E
=
Input dd
fu. Diff w.r.t. d
=10.I-w
PMG
0:
En-w
= w
-
x
=
E
y =
(e) i TR
=
- TC
P.y-w.c
=10
It-(1)
:10
-
=10 (5):25
PMG
50-25
=
25
=
98993/2641
PMG
1. Dec Notel
-
-
AboutIMG
0.(a)y x I
= +
P6
=
MP(n)
b*
=
=
=
ns+
API) =v's+ I
=
PMG
(b) XMP (n). P
=
MP()
x
~AP(U1 PxAPa)
=
b-
-
(= !] =s(us+]
=
Y5 2
+
6x
=
2
+
(C) VMP=w
4x5 +
2 = W
Yut W-2
=
=
ns=
e
x (wE2)"
=
Inputdd fu
05:
-
*
Concept of Multiple Outputs One
Inputonly -> x
Two outputs ->
y, Yu
X (y,,y2) f(u)
=
given
1 "(y,y) X
PMG
x =
j-1
y y2
Max Its:4y,+Ryz-wise
i P,y,
=
+
Pcyc- W.
I"(y,,y2)
-1, -wy, 28(y,,y)] : is-way. (8" (4,,y1]
0 12 -
MC2
P,
=
MC,
PMG
0 = -
P2 MC2
y,
=
P, MC,
=
y,
- 10 unit
of x
X W
--
25000
->
Profits,
# i TR-TC =
Letg f(1)
=
it P.q-W.
=
4
P.f(1)
=
-
w. 1
l
Diff. w.r.t.
= It -
PMG
w
0 = P. MP -
w
YMP w
= First order Condition
I=P-w
Diff w.r.t. I
again
: "(slope of MP 0
-
=9 MP
Slope of
*
For Returne
Man.
of us,
co 1.0 Slope of MP <0 i.2
Dive.
>
