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Review Question Set #4 : Answer Keys
1. Define G(x, y) = 2x2 + 6xy + y2 18 then we have G(x, y) = 0.
Applying the implicitfunction theorem, we have
y
x
(1,2)
= GxGy
= 4x + 6y2y + 6x
(1,2)
= 85
2. Substitute f(Y), g(Y) for C, M and define H(X , I , Y ) Yf(Y)
+ g(Y)IX = 0.Then
Y
I= HI
HY= 1
1 f(Y) + g(Y)
When investment increases, output increases much if f(Y)
(sensitivity of consumption
to output) is big, and g(Y) (sensitivity of import to output) is
small.
3. (a) f(x) = 2(b) f(x) = 3x2 3(c) f(x) = 1 1/x2
4. (a) f(x) = 3x2 + 64x 6, f(x) = 6x + 64.
(b)x =
64 642 4 3 62 3 .
f is increasing when f 0. It happens when x is outside of two
solutions here.
(c) Inflection point is where f = 0, In other words, where f
changes signs. Then
the curve becomes either convex from concave or concave from
convex.
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Inflection point is x = 32/3.
5. (a) Revenue : p x. Cost : w xProfit () : p
x wx.
(b) FOC : 12px(1/2) w. Value of marginal product = input factor
price (marginal
cost of adding one unit of input).
(c) SOC : 14px(3/2) < 0
(d) () : px2 wx. FOC : 2px w = 0. SOC : 2p > 0. The
stationary point is notprofit maximizing. Because production
function has increasing return to scale, a
firm can increase profit whenever level of production
increases.
6. Since f is concave, we have f(x + (1 )y) f(x) + (1 )f(y)
(*).By definition, g(x + (1 )y) = af(x + (1 )y) + b.g is concave
if
af(x + (1 )y) + b af(x) + (1 )af(y) + b
Note this is just equivalent to (*) whenever a 0.
7. (a) Define y =T
t=1 yt. Utility from constant consumption :
1
T
u(1
T
y) +1
T
u(1
T
y) + ... +1
T
u(1
T
y) = u(1
T
y) (
)
Here Jensens inequality is just the property of concave function
(u(ax+(1a)y) au(x)+(1a)u(y)). For any consumption stream c1 = 1y,
c2 = 2y, ..., cT = Ty,1 + 2 + ... + T = 1, we have from Jensens
inequality
u(1
Ty) = u(
1
T0y + ... +
1
TTy) 1
Tu(0y) + ... +
1
Tu(Ty)
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Thus lifetime utility is maximized when the consumption is
constant for every
period. Formally, Jensens inequality states that for a concave
function ,
(
aixiaj
)
ai(xi)aj
.
(b) We can set up a Lagrangian as
L :
T
t=1
(1 + r)tu(ct) + [
T
t=1
(1 + r)t(yt ct)]
FOC : Lct : u(ct) = for every t = 1,...,T, Implying
c1 = c2 = ... = cT =
Tt=1(1 + r)
tytTt=1(1 + r)
t
8. h(x) = min{f(x), g(x)}, f and g are concave functions. Then
we have
f(x + (1 )y) f(x) + (1 )f(y)
g(x + (1 )y) g(x) + (1 )g(y)
h(x + (1 )y) = min{f(x + (1 )y), g(x + (1 )y)}
min{f(x), g(x)} + (1 )min{f(y), g(y)} = h(x) + (1 )h(y)
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