Homework Assignment #4
13.15 Prove that a linear function from Rk to Rm sends a line in Rk to a point or a line in Rm .
Answer: A line can be written as ` = {x0 + tv : t ∈ R}. Let L be a linear function from Rk → Rm .
Then L(x0 + tv) = Lx0 + tLv. If Lv = 0, the line ` is mapped to a single point. Otherwise, it is mapped
to {Lx0 + tLv : t ∈ R}, which is a line in Rm .
3/2
14.6 Consider the constant elasticity demand function Q = 6p−2
where Q is the demand for good 1 and
1 p2
pi is the price of good i = 1, 2. Suppose current prices are p1 = 6 and p2 = 9.
a) What is the current demand for Q?
b) Use differentials to estimate the change in demand as p1 increases by 0.25 and p2 decreases by 0.5.
c) Similarly, estimate the change in demand when both prices increase by 0.2.
d) Estimate the totals demands for situations b and c and compare your estimates with the actual
demands.
Answer:
a) Demand is Q = 4.5.
3/2
1/2
−2
b) Now dQ = −12p−3
1 p2 dp1 +9p1 p2 dp2 . Evaluating at (p1 , p2 ) = (6, 9) we obtain dQ = −(3/2)dp1 +
(3/4)dp2 . Setting dp1 = 0.25 and dp2 = −0.5, we find dQ = −3/8 − 3/8 = −3/4 = −0.75.
c) Here dQ = −.3 + 0.15 = −0.15.
d) The actual values to 3 decimal places are Qb = 3.806 and Qc = 4.356, so ∆Qb = −0.694 vs. −0.75
and ∆Qc = −0.144 vs. −0.15.
14.27 Consider the production function Q = K 3/4 L3/4 . Show that marginal productivity of each factor is
diminishing. Show, however, that for any strictly positive input combination, if the input combination
is doubled, then output more than doubles.
∂Q
3/4 −1/4
= (3/4)K −1/4 L3/4 , which is a diminishing function of K and ∂Q
L
,
Answer: Here ∂K
∂L = (3/4)K
3/2
which is a decreasing function of L. Finally, Q(2K, 2L) = 2 Q(K, L) > 2Q(K, L).
14.28 The goal of this exercise is to examine a C 1 function for which the conclusion of Theoorem 14.5 fails—the
cross partials are not equal. Let
(
0
if (x, y) = (0, 0),
f (x, y) =
x3 y−xy 3
otherwise.
x2 +y 2
a) Prove that f is zero along the x-axis and the y-axis. Conclude that (∂f /∂x)(0, 0) and (∂f /∂y(0, 0)
are both 0.
b) Compute ∂f /∂x and ∂f /∂y for (x, y) 6= (0, 0).
c) Conclude that (∂f /∂x)(0, y) = −y and (∂f /∂x)(x, 0) = x.
d) Show that
∂
∂2f
(0, 0) =
∂y∂x
∂y
∂2f
∂
(0, 0) =
∂x∂y
∂x
∂f
∂x
(0, 0) = lim
∂f
∂x (0, y)
−
y
∂f
∂x (0, 0)
∂f
∂y (x, 0)
−
∂f
∂y (0, 0)
y→0
= −1.
e) Show that
∂f
∂y
(0, 0) = lim
x→0
x
= +1
HOMEWORK ASSIGNMENT #4
Page 2
and conclude that the mixed partials are not equal at (0, 0).
f) Compute (∂ 2 f /∂x∂y)(x, y) for all (x, y) other than (0, 0).
g) Use f to show that (∂ 2 f /∂x∂y)(x, x) = 0 for x > 0.
h) Compare e and g to show that (∂ 2 f /∂x∂y)(x, y) is discontinuous at the origin. Therefore, f is not
C 2 and the hypotheses of Theorem 14.5 do not hold.
Answer:
a) For y 6= 0, f (0, y) = 0/y 2 = 0 and for x 6= 0, f (x, 0) = 0/x2 = 0. Finally, f (0, 0) = 0 by definition.
b) Now
x3 y − xy 3
x4 y + 4x2 y 3 − y 5
3x2 y − y 3
∂f
−
2x
=
= 2
∂x
x + y2
(x2 + y 2 )2
(x2 + y 2 )2
and
∂f
x3 − 3xy 2
x3 y − xy 3
x5 − 4x3 y 2 − xy 4
=
− 2y 2
=
.
2
2
2
2
∂y
x +y
(x + y )
(x2 + y 2 )2
5
4
c) For y 6= 0, (∂f /∂x)(0, y) = −y 5 /y4 = −y. For x 6= 0, ( ∂f
∂y )(x, 0) = x /x = x.
d) The expression is just the difference quotient defining ∂ 2 f /∂x∂y. It is true because
∂f
∂y (0, y)
−
∂f
∂x (0, 0)
y
=
−y
= −1.
y
e) The expression is just the difference quotient defining ∂ 2 f /∂y∂x. It is true because
∂f
∂y (x, 0)
−
∂f
∂y (0, 0)
x
=
x
= +1.
x
The order of differentiation makes a difference here.
f) Now
∂2f
∂
=
∂x∂y
∂x
∂f
∂y
=
x6 + 9x4 y 2 − 9x2 y 4 − y 6
.
(x2 + y 2 )3
g) Setting y = x 6= 0, we find
∂2f
0
= 6 6 = 0.
∂x∂y
2 x
h) Now
∂2f
∂2f
(x, x) = 0 6= +1 =
(0, 0)
x→0 ∂x∂y
∂x∂y
lim
so the second derivative is not continuous at (0, 0).
15.8 Consider the equation x3 + 3y 2 + 4xz 2 − 3z 2 y = 1. Does this equation define z as a function of x and y:
a) In a neighborhood of x = 1, y = 1?
b) In a neighborhood of x = 1, y = 0?
c) In a neighborhood of x = 0.5, y = 0? If so, compute ∂z/∂x and ∂z/∂y at this point.
Answer:
a) When x = 1 and y = 1, 4 + z 2 = 1, which has no solution z.
HOMEWORK ASSIGNMENT #4
Page 3
2
b) When x = 1 and y = 0, 1 + 4z = 1, implying z = 0. Now dz f = (8x − 6y)z = 0. As this is zero,
the implicit function theorem does not apply. In fact, at (1 + , 0), there are no solutions for small
> 0.
√
√
c) Here z = ± 7/4. Since ∂f /∂z = 4z = ± 7, the implicit function theorem yields two solutions near
√
(x, y) = (0.5, 0). The derivatives are ∂z/∂x = −∂f /∂x/4z = ∓5 7/14 and ∂z/∂y = −∂f /∂y/4z =
√
±3 7/16.
15.13 A firm uses x hours of unskilled labor and y hours of skilled labor each day to produce Q(x, y) =
60x2/3 y 1/3 units of output per day. It currently employs 64 hours of unskilled labor and 27 hours of
skilled labor.
a) What is its current output?
b) In what direction (expressed as a unit vector) should it change (x, y) if it wants to increase output
most rapidly?
c) The firm is planning to hire an additional hour and a half of skilled labor. Use calculus to estimate
the corresponding change in unskilled labor that would keep its output at its current level.
Answer:
a) Its current output is Q(64, 27) = 60(24 )3 = 2880.
b) Here dQT = Q(2/3x, 1/3y)T = 2880(1/96, 1/81)T = (30, 35 59 )T is the direction of fastest increase.
√
To express that as a unit vector, we divide it by k(30, 35 95 )T k = 10 1753/9 ≈ 46.52, obtaining
1753−1/2 (27, 32)T ≈ (0.58, 0.69)T
c) Using the implicit function theorem, we find we may treat x as a function of y because ∂Q/∂x =
30 6= 0. Then dx/dy = −(35 59 )/30 = −32/27. When we increase y by 1.5 we must decrease x by
(−32/27)(1.5) = −16/9 in order to maintain output at its current level.